Exercise 8.1
Q.1: Expand the Expression (1 – 3x)5
Sol.
By using Binomial Theorem:
Since, (1 – x)n = [ nC0 ] + [ nC1 × (-x) ] + [ nC2 × (-x)2 ] + [ nC3 × (-x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (-x)n ]
Therefore, (1 – 3x)5= [ 5C0 ] – [ 5C1 × (3x) ] + [ 5C2 × (3x)2 ] – [ 5C3 × (3x)3 ] + [ 5C4 × (3x)4 ] – [ 5C5 × (3x)5 ] ⇒ 1 – [ 5 × (3x) ] + [ 10 × (9x2) ] – [ 10 × (27x3) ] + [ 5 × (81x4) ] – [ 1 × (243x5) ] ⇒ 1 – 15x + 90x2 – 270x3 + 405x4 – 243x5
Therefore, (1 – 3x)5 = 1 – 15x + 90x2 – 270x3 + 405x4 – 243x5
Q.2: Expand the Expression (5x−x5)5
Sol.
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]
Therefore, (5x−x5)5:
[5C0 × (5x)5 ] – [5C1 × (5x)4 × (x5)1 ] + [5C2 × (5x)3 × (x5)2 ] – [ 5C3 × (5x)2 × (x5)3 ] + [5C4 × (5x)1 × (x5)4 ] – [5C5 × (x5)5] ⇒[1×3125x5]−[5×625x4×x5]+[10×125x3×x225]−[10×25x2×x3125]+[5×5x×x4625]−[1×x53125] ⇒[3125x5]−[625x3]+[50x]−2x+[3x325]−[x53125]
Therefore, (5x−x5)5:
[x53125]+[3x325]−2x+[50x]−[625x3]+[3125x5]
Q.3: Expand the Expression (3x – 2)6
Sol.
By using Binomial Theorem:
[x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]
Therefore, (3x – 2)6 = [ 6C0 × (3x)6 ] – [ 6C1 × (3x)5 × (2) ] + [ 6C2 × (3x)4 × (2)2 ] – [ 6C3 × (3x)3 × (2)3 ] + [ 6C4 × (3x)2 × (2)4 ] – [ 6C5 × (3x)1 × (2)5 ] + [ 6C6 × (2)6 ] ⇒ [1 × (729x6)] – [6 × (243x5) × 2] + [15 × (81x4) × 4] – [20 × (27x3) × 8 ] + [15 × (9x2) × 16] – [6 × (3x) × 32] + [1 × 64] ⇒ 729x6 – 2916x5 + 4860x4 – 4320x3 + 2160x2 – 576x + 64
Therefore, (3x – 2)6 = 729x6 – 2916x5 + 4860x4 – 4320x3 + 2160x2 – 576x + 64
Q.4: Expand the Expression (4x+x3)5
Sol.
By using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, (4x+x3)5:
[ 5C0 × (4x)5 ] + [ 5C1 × (4x)4 × (x3)1 ] + [ 5C2 × (4x)3 × (x3)2 ] + [ 5C3 × (4x)2 × (x3)3 ] + [ 5C4 × (4x)1 × (x3)4 ] + [ 5C5 × (x3)5] ⇒[1×1024x5]+[5×256x4×x3]+[10×64x3×x29]+[10×16x2×x327]+[5×4x×x481]+[1×x5243] ⇒[1024x5]+[12803x3]+[6409x]+[160x27]+[20x381]+[x5243]
Therefore, (4x+x3)5:
⇒[x5243]+[20x381]+[160x27]+[6409x]+[12803x3]+[1024x5]
Q.5: Expand the Expression (1x−x2)6
Sol.
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]
Therefore, (1x−x2)6:
[ 6C0 × (1x)6 ] – [ 6C1 × (1x)5 × (x2)1 ] + [ 6C2 × (1x)4 × (x2)2 ] – [ 6C3 × (1x)3 × (x2)3 ] + [ 6C4 × (1x)2 × (x2)4 ] – [ 6C5 × (1x)1 × (x2)5 ] + [ 6C6 ×(x2)6] ⇒[1×1x6]−[6×1x5×x2]+[15×1x4×x24]−[20×1x3×x38]+[15×1x2×x416]−[6×1x×x532]+[1×x664] ⇒[1x6]−[3x4]+[154x2]−[52]+[15x216]−[3x416]+[x664]
Therefore, (1x−x2)6:
⇒[x664]−[3x416]+[15x216]−[52]+[154x2]−[3x4]+[1x6]
Q.6: By using Binomial Theorem, Evaluate (98)4
Sol.
(98)4 = (100 – 2)4
Now, by using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]
Therefore, (100 – 2)4 = [ 4C0 × (100)4 ] – [ 4C1 × (100)3 × (2) ] + [ 4C2 × (100)2 × (2)2 ] – [ 4C3 × 100 × (2)3 ] + [4C4 × (2)4] ⇒ (1 × 100000000) – (4 × 1000000 × 2) + (6 × 10000 × 4) – (4 × 100 × 8 ) + (1 × 16)
⇒ 100000000 – 8000000 + 240000 – 3200 + 16
Therefore, (98)4 = (100 – 2)4 = 92236816 = 9.2236816 × 107
Q.7: By using Binomial Theorem, Evaluate (105)5
Sol.
(105)5 = (100 + 5)5
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, (100 + 5)5 = [ 5C0 × (100)5 ] + [ 5C1 × (100)4 × (5) ] + [ 5C2 × (100)3 × (5)2 ] + [ 5C3 × (100)2 × (5)3 ] + [ 5C4 × (100) × (5)4 ] + [ 5C5 × (5)5] ⇒ (1 × 10000000000) + (5 × 100000000 × 5) + (10 × 1000000 × 52 ) + (10 × 10000 × 53 ) + (5 × 100 × 54 ) + ( 1 × 55 )
⇒ 10000000000 + 2500000000 + 250000000 + 12500000 + 312500 + 3125
Therefore, (105)5 = (100 + 5)5 = 12762815630 = 1.276281563 × 1010
Q.8: By using Binomial Theorem, Evaluate (104)4
Sol.
(104)4 = (100 + 4)4
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, (100 + 4)4 = [4C0 × (100)4 ] + [ 4C1 × (100)3 × (4) ] + [4C2 × (100)2 × (4)2] + [4C3 × 100 × (4)3] + [4C4 × (4)4 ] ⇒ (1 × 100000000) + (4 × 1000000 × 4) + (6 × 10000 × 16) + (4 × 100 × 64 ) + (1 × 256)
⇒ 100000000 + 16000000 + 960000 + 25600 + 256
Therefore, (104)4 = (100 + 4)4 = 116985856 = 1.16985856 × 108
Q.9: By using Binomial Theorem, Evaluate (95)5
Sol.
(95)5 = (100 – 5)5
Now, by using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, (100 – 5)5 = [ 5C0 × (100)5 ] – [ 5C1 × (100)4 × (5) ] + [ 5C2 × (100)3 × (5)2 ] – [ 5C3 × (100)2 × (5)3 ] + [ 5C4 × (100) × (5)4 ] – [ 5C5 × (5)5 ] ⇒ (1 × 10000000000) – (5 × 100000000 × 5) + (10 × 1000000 × 52 ) – (10 × 10000 × 53 ) + (5 × 100 × 54 ) – ( 1 × 55 )
⇒ 10000000000 – 2500000000 + 250000000 – 12500000 + 312500 – 3125
Therefore, (95)5 = (100 – 5)5 = 7737809375 = 7.737809375 × 109
Q.10: By using Binomial Theorem, determine which number is greater (1.1)10000 or 1000.
Sol.
Now, (1.1)10000 = (1 + 0.1)10000
Now, on applying Binomial Theorem, the first few terms in the expansion of (1 + 0.1)10000 are:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, (1 + 0.1)10000 = [ 10000C0 × (1)10000 ] + [ 10000C1 × (1)9999 × 0.1 ] + other positive terms
⇒ [1 × 1] + [10000 × 1 × 0.1] + other positive terms
⇒ 1 + 1000 + other positive terms > 1000
Therefore, (1.1)10000 is greater than 1000.
Q.11: Find (a + b)5 – (a – b)5 . Hence evaluate (5–√+3–√)5−(5–√−3–√)5
Sol.
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, (a + b)5= [ 5C0 × (a5) ] + [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] + [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] + [ 5C5 × b5 ]
And, [a – b]5 = [ 5C0 × (a5) ] – [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] – [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] – [ 5C5 × b5 ]
Therefore, (a + b)5 – (a – b)5 = [ 5C0 a5 + 5C1 a4 b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5 ] – [ 5C0 a5 – 5C1 a4b + 5C2 a3b2 – 5C3 a2b3 + 5C4 ab4 – 5C5 b5 ] ⇒ [ 5C0 a5 + 5C1 a4 b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5 – 5C0 a5 + 5C1 a4b – 5C2 a3b2 + 5C3 a2b3 – 5C4 ab4 + 5C5 b5 ] ⇒ 2[ 5C1 a4b + 5C3 a2b3 + 5C5 b5 ] = 2[ 5a4b + 10a2b3 + b5 ]
Therefore, (a + b)5 – (a – b)5 = 2b[ 5a4 + 10a2 b2 + b4 ] . . . . . . (1)
Now, on substituting a = 5–√ and b = 3–√ in equation (1) we will get:
(5–√+3–√)5−(5–√−3–√)5=23–√×[5(5–√)4+10(5–√)2×(3–√)2+(3–√)4]
⇒23–√[125+150+9]=5683–√
Therefore, (5–√+3–√)5−(5–√−3–√)5=5683–√
Q.12: Find (1 + x)5 – (1 – x)5 . Hence evaluate (1+7–√)5−(1−7–√)5
Sol.
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . + [ nCn × yn ]
And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, (1 + x)5= [ 5C0 × (15) ] + [ 5C1 × (14) × (x) ] + [ 5C2 × (13) × (x)2 ] + [ 5C3 × (12) × (x)3 ] + [ 5C4 × (11) × (x)4 ] + [ 5C5 × (x)5 ]
And, [1 – x]5 = [ 5C0 × (15) ] – [ 5C1 × (14) × (x) ] + [ 5C2 × (13) × (x)2 ] – [ 5C3 × (12) × (x)3 ] + [ 5C4 × (11) × (x)4 ] – [ 5C5 × (x)5 ]
Therefore, (1 + x)5 – (1 – x)5 = [ 5C0 + 5C1 x + 5C2 x2 + 5C3 x3 + 5C4 x4 + 5C5 x5 ] – [ 5C0 – 5C1 x + 5C2 x2 – 5C3 x3 + 5C4 x4 – 5C5 x5 ] ⇒ [ 5C0 + 5C1 x + 5C2 x2 + 5C3 x3 + 5C4 x4 + 5C5 x5 – 5C0 + 5C1 x – 5C2 x2 + 5C3 x3 – 5C4 x4 + 5C5 x5 ] ⇒ 2[ 5C1x + 5C3 x3 + 5C5 x5 ] = 2[ 5x + 10x3 + x5 ]
Therefore, (1 + x)5 – (1 – x)5 = 2x[ 5 + 10x2 + x4 ] . . . . . . (1)
Now, on substituting x = 7–√ in equation (1) we will get:
⇒(1+7–√)7−(1−7–√)5=27–√[5+10(7–√)2+(7–√)4] ⇒27–√[5+70+49]=2487–√
Therefore, (1+7–√)5−(1−7–√)5=2487–√
Q.13 Show that 7n+1 – 6n – 7 is divisible by 36, where n is the positive integer.
Sol.
Equation 7n+1 – 6n – 7 will be divisible by 36 if, 7n+1 – 6n – 7 = 36p [where p is any natural number]
Now, by Binomial Theorem:
Since, [1 + x]m = [ mC0 ] + [ mC1 . x ] + [ mC2 . x2 ] + [ mC3 . x3 ] + . . . . . . . . . . . . + [ mCm . xm ]
Therefore, for x = 6 and m = n + 1 the equation becomes:
[1 + 6]n+1 = [ n+1C0 ] + [ n+1C1 × 6 ] + [ n+1C2 × (6)2 ] + [ n+1C3 × (6)3 ] + . . . . . . . . . . . . . + [ n+1Cn+1 × (6)n+1 ] ⇒ [7]n + 1 = 1 + [(n + 1) × 6] + 62 [ n+1C2 + n+1C3 × (6) + . . . . . . . . . + n+1Cn + 1 × (6)n–1 ] ⇒ [7]n + 1 = 1 + 6n + 6 + 36 [ n+1C2 + n+1C3 × (6) + . . . . . . . . . + n + 1Cn + 1 × (6)n – 1 ]
⇒ (7)n + 1 – 6 n – 7 = 36p
Where p is any natural number and p = [n + 1C2 + n + 1C3× (6) + . . . . . . . . . . . . . . . . . . + n + 1Cn + 1 × (6)n – 1]
Therefore, 7n+1 – 6n – 7 is divisible by 36 [where n is a positive integer]
Q.14 Prove that ∑nr=0 2r× nCr = 3n
Sol.
By using Binomial Theorem:
Since, (a + b)n = ∑nr=0 nCr (a)n – r × (b)r
Therefore, on substituting the value of a = 1 and b = 2 in the above equation we will get:
(1 + 2)n = ∑nr=0 nCr (1)n – r × (2)r
⇒ 3n = ∑nr=0 nCr× (2)r
Therefore, ∑nr=0 2r× nCr = 3n
Q.15 Show that 5n – 4n always leaves remainder 5 when divided by 16, where n is the positive integer.
Sol.
Equation 5n – 4n will leave remainder 5 when divided by 16 if, 5n – 4n = 16p + 5 [where p is any natural number]
Now, by Binomial Theorem:
Since, [1 + x]n = [ nC0 ] + [ nC1 . x ] + [ nC2 . x2 ] + [ nC3 . x3 ] + . . . . . . . . . . . . . . . . . . + [ nCn . xn ]
Therefore, for x = 4 the equation becomes:
[1 + 4]n = [ nC0 ] + [ nC1 × 4 ] + [ nC2 × (4)2 ] + [ nC3 × (4)3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × (4)n ] ⇒ [5]n = 1 + [(n + 1) × 4] + 42 [ nC2 + nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n ] ⇒ [5]n = 1 + 4n + 4 + 16 [ nC2 + nC3 × (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n–1 ]
⇒ 5n – 4n – 5 = 16p
Where p is any natural number and p = [ nC2 + nC3× (4) + . . . . . . . . . . . . . . . . . . + nCn × (4)n–1]
i.e. 5n – 4n = 16 + 5
Therefore, 5n – 4n will leave remainder 5 when divided by 16, where n is any natural number.
Q.16: Find (a + b)6 – (a – b)6 . Hence evaluate: (2–√+3–√)6−(2–√−3–√)6
Sol.
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, [a + b]6= [ 6C0 × (a6) ] + [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] + [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] + [ 6C5 × a × b5 ] + [6C6 × b6]
And, [a – b]6 = [ 6C0 × (a6) ] – [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] – [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] – [ 6C5 × a × b5 ] + [6C6 × b6]
Therefore, (a + b)6 – (a – b)6 = [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b6] – [ 6C0 a6 – 6C1 a5b + 6C2 a4b2 – 6C3 a3b3 + 6C4 a2b4 – 6C5 ab5 + 6C6 b6 ] ⇒ [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b6 – 6C0 a6 + 6C1 a5b – 6C2 a4b2 + 6C3 a3b3 – 6C4 a2b4 + 6C5 ab5 – 6C6 b6 ] ⇒ 2[ 6C1 a5b + 6C3 a3b3 + 6C5 ab5 ] = 2[ 6a5b + 20a3b3 + 6ab5 ]
Therefore, (a + b)6 – (a – b)6 = 4ab[ 3a4 + 10a2 b2 + 3b4 ] . . . (1)
Now, on substituting a = 2–√ and b = 3–√ in equation (1) we will get:
(2–√+3–√)6−(2–√−3–√)6=4(2–√.3–√)[3(2–√)4+10(2–√)2×(3–√)2+3(3–√)4] ⇒46–√[12+60+27]=3966–√
Therefore, (2–√+3–√)6−(2–√−3–√)6=3966–√
Q.17: By using Binomial Theorem, Evaluate (91)4
Sol.
(91)4 = (100 – 9)4
Now, by using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]
Therefore, (100 – 9)4 = [ 4C0 × (100)4 ] – [ 4C1 × (100)3 × (9) ] + [ 4C2 × (100)2 × (9)2 ] – [ 4C3 × 100 × (9)3 ] + [4C4 × (9)4] ⇒ (1 × 100000000) – (4 × 1000000 × 9) + (6 × 10000 × 81) – (4 × 100 × 729) + (1 × 6561)
⇒ 100000000 – 36000000 + 4860000 – 291600 + 6561
Therefore, (91)4 = (100 – 9)4 = 68574961 = 6.8574961 × 107
Q.18: By using Binomial Theorem, Evaluate (107)5
Sol.
(107)5 = (100 + 7)5
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, (100 + 7)5 = [ 5C0 × (100)5 ] + [ 5C1 × (100)4 × (7) ] + [ 5C2 × (100)3 × (7)2 ] + [ 5C3 × (100)2 × (7)3 ] + [ 5C4 × (100) × (7)4 ] + [ 5C5 × (7)5] ⇒ (1 × 10000000000) + (5 × 100000000 × 7) + (10 × 1000000 × 72 ) + (10 × 10000 × 73 ) + (5 × 100 × 74 ) + ( 1 × 75 )
⇒ 10000000000 + 3500000000 + 490000000 + 34300000 + 1200500 + 16807
Therefore, (107)5 = (100 + 7)5 = 1402551731 = 1.402551731 × 1010
Q.19: Expand the Expression (32x−x5)6
Sol.
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]
Therefore, (32x−x5)6:
[6C0 × (32x)6 ] – [6C1 × (32x)5 × (x5)1 ] + [6C2 × (32x)4 × (x5)2 ] – [6C3 × (32x)3 × (x5)3 ] + [6C4 × (32x)2 × (x5)4 ] – [6C5 × (32x)1 × (x5)5 ] + [6C6 ×(x5)6] ⇒[1×72964×x6]−[6×24332×x5×x5]+[15×8116×x4×x225]−[20×278x3×x3125]+[15×94x2×x4625]−[6×32x×x53125]+[1×x615625] ⇒[72964x6]−[72980x4]+[24380x2]−[2750]+[27x2500]−[9x43125]+[x615625]
Therefore, (32x−x5)6:
⇒[x615625]−[9x43125]+[27x2500]−[2750]+[24380x2]−[72980x4]+[72964x6]
Q.20: Expand the Expression (5x – 3)6
Sol.
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . +(-1)n [ nCn × yn ]
Therefore, (5x – 3)6 = [ 6C0 × (5x)6 ] – [ 6C1 × (5x)5 × (3) ] + [ 6C2 × (5x)4 × (3)2 ] – [ 6C3 × (5x)3 × (3)3 ] + [ 6C4 × (5x)2 × (3)4 ] – [ 6C5 × (5x)1 × (3)5 ] + [ 6C6 × (3)6 ] ⇒ [1 × (15625 x6)] – [6 × (3125 x5) × 3] + [15 × (625 x4) × 9] – [20 × (125 x3) × 27 ] + [15 × (25 x2) × 81] – [6 × (5x) × 243] + [1 × 729] ⇒ 15625 x6 – 56250 x5 + 84375 x4 – 67500 x3 + 30375 x2 – 7290 x + 729
Therefore, (5x – 3)6 = 15625 x6 – 56250 x5 + 84375 x4 – 67500 x3 + 30375 x2 – 7290 x + 729
Exercise 8.2
Q.1: Find the Coefficient of x6 in the expansion of (x + 2)9
Sol.
Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Therefore, on substituting n = 9, a = x and b = 2 in the above expression we will get:
Tr + 1 = 9Cr × (x)9 – r× 2r . . . . . . . (1)
Now, on comparing the coefficient of x in equation (1) with x6, we will get:
i.e. (x)9 – r = x6
(or) 9 – r = 6
Therefore, r = 3
Now, on substituting the value of r in equation (1) we will get:
T4 = 9C3 × (x)9 – 3 × 23
⇒T4=9!3!6!×x6×8 ⇒T4=9×8×7×6!3×2×1×6!×8x6=672x6
Therefore, the Coefficient of x6 in the expansion of (x + 2)9 = 672
Q.2: Find the Coefficient of a6 b8 in the expansion of (2a – 3b)14
Sol.
Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r× yr
Therefore, on substituting n = 14, x = 2a and y = (-3b) in the above expression we will get:
Tr + 1 = 14Cr × (2a)14 – r × (-3b)r
i.e. Tr + 1 = [14Cr × (2)14 – r (-3)r ] (a)14 – r× (b)r . . . . . . . . . . (1)
Now, on comparing the coefficients of a and b in equation (1) with a6 b8, we will get:
i.e. a6 b8 = a14 – r br
Therefore, r = 8
Now, on substituting the value of ‘r’ in equation (1) we will get:
T9 = [ 14C8 × (2)14 – 8 (-3)8 ] (a)14 – 8 × (b)8
i.e. T9 = [ 14C8 × (2)6 (-3)8 ] a6 b8
⇒T9=14!8!6!×64×729×a6b8 ⇒T9=14×13×12×11×10×9×8!6×5×4×3×2×1×8!×64×729×a6b8 ⇒T9=3003×64×729×a6b8=140107968a6b8
Therefore, the Coefficient of a6 b8 in the expansion of (2a – 3b)14 = 140107968
Q.3: Write the general term in the expansion of (x3 – y2)5
Sol.
The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Now, on substituting n = 5, a = x3 and b = y2 we will get:
Tr+1 = 5Cr × (x3)5 – r× (y2)r
Therefore, the general term in the expansion of (x3 – y2)5:
Tr+1 = 5Cr × (x)15 – 3 r× (y)2 r
Q.4: Write the general term in the expansion of (x4 – xy2)9
Sol.
The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Now, on substituting n = 9, a = x4 and b = xy2 we will get:
Tr+1 = 9Cr × (x4)9 – r× (xy2)r
⇒ Tr+1 = 9Cr × (x)36 – 4 r × (x)2 r × (y)2 r
⇒ Tr+1 = 9Cr × (x)36 – 4 r + 2 r × (y)2 r
⇒ Tr+1 = 9Cr × (x)36 – 2 r × (y)2 r
Therefore, the general term in the expansion of (x4 – xy2)9:
Tr+1 = 9Cr × (x)36 – 2 r× (y)2r
Q.5: Find the 4th term in the expansion of (2x + 3y)10
Sol.
Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Therefore, from the above equation the value of r should be 3; for finding out the values of 4th Term (T4)
Therefore, T3 + 1 = nC3 × (a)n – 3× b3
Now, on substituting n =10, a = 2x and b = 3y we will get:
⇒ T3 + 1 = 10C3 × (2x)10 – 3 × (3y)3
⇒ T4 = 10C3 × (2x)7 × (3y)3
⇒T4=10!3!7!×128x7×27y3 ⇒T4=10×9×8×7!3×2×1×7!×3456x7y3 ⇒T4=120×3456x7y3=414720x7y3
Therefore, 4th term in the expansion of (2x + 3y)10 = 414720 x7y3
Q.6 Find the 11th term in the expansion of (8x−12x√)15
Sol.
Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Therefore, from the above equation the value of r should be 10 for finding out the values of 11th Term (T11)
Therefore, T10 + 1 = nC10 × (a)n – 10× b10
Now, on substituting n =15, a = 8x and b = (−12x√) we will get:
⇒ T10 + 1 = 15C10 × (8x)15 – 10 × (−12x√)10 ⇒ T11 = 15C10 × (8x)5 × (−12x√)10 ⇒T11=15!10!5!×8×8×8×8×8×x5×1210×(1x√)10 ⇒T11=15×14×13×12×11×10!5×4×3×2×1×10!×32×x5×1x5 ⇒T11=3003×32=96096
Therefore, 11th term in the expansion of (8x−12x√)15= 96096
Q.7: In the expansion of (5−x410)9, Find the middle terms.
Sol.
Here, n = 9
When n is odd then the middle terms in the expansion of (a + b)n are given by:
middle terms = (n+12)thtermand(n+12+1)thterm
Therefore, the middle terms in the expansion of (5−x410)9are:
⇒(9+12)thtermand(9+12+1)thterm
⇒ 5th term and 6th term
Now, 5th and 6th terms in the expansion of (5−x410)9 are:
T5 = T4 + 1 = 9C4 × (5)9 – 4 × (−x410)4 ⇒ T5 = 9C4 × (5)5 × (−x410)4 ⇒T5=9!4!5!×(5)5×(−x410)4 ⇒T5=9×8×7×6×5!4×3×2×1×5!×5×5×5×5×5×1104×x16 ⇒T5=126×516x16=3158x16
Therefore, 5th term =3158x16
Now, T6 = T5 + 1 = 9C5 × (5)9 – 5 × (−x410)5 ⇒ T6 = 9C5 × (5)4 × (−x410)4 ⇒T6=9!5!4!×(5)4×(−x410)5 ⇒T6=9×8×7×6×5!5!×(4×3×2×1)×5×5×5×5×−1105×x20 ⇒T6=−126×1320x20=−63160x20
Therefore, 6th term =63160x20
Therefore, 5th and 6th terms are the middle terms in the expansion of (5−x410)9
And also, 5th term =3158x16 and 6th term =−63160x20
Q.8: In the expansion of (x2+8y)10, Find the middle term.
Sol.
Here, n = 10
When n is even, the middle term in the expansion of (a + b)n is given by:
(n2+1)thterm
Therefore, the middle term in the expansion of (x2+8y)10 is:
⇒(102+1)thterm = 6th term
Now, 6th term in the expansion of (x2+8y)10 is:
T6 = T5 + 1 = 10C5 × (x2)10−5 × (8y)5
⇒ T6 = 10C5 × (x2)5 × (8y)5
⇒T6=10!5!5!×(x2)5×(8y)5 ⇒T6=10×9×8×7×6×5!5×4×3×2×1×5!×x532×8×8×8×8×8×y5 ⇒T6=252×1024x5y5=258048x5y5
Therefore, 6th term = 258048 x5 y5
Hence, the middle term in the expansion of (x2+8y)10 = 258048 x5 y5
Q.9: Prove that the coefficients of pa and pb are equal in the expansion of (1 + p)a + b
Sol.
Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r× (y)r
Now, on substituting n = (a + b), x =1 and y = p, we will get:
Tr + 1 = (a + b)Cr × (1)(a + b) – r × (p)r
⇒ Tr + 1 = (a + b)Cr× (p)r . . . . . . . . . . . . (1)
Now, on comparing the coefficient of p in equation (1) with pa, we will get r = a
Therefore, from equation (1):
T a + 1 = (a + b)Ca × (p)a
⇒Ta+1=(a+b)!a!(a+b−a)!×pa ⇒Ta+1=(a+b)!a!b!×pa
Therefore, the coefficient of (p)a = (a+b)!a!b! . . . . . . . . . . (2)
Now, on comparing the coefficient of p in equation (1) with pb, we will get r = b
Therefore, from equation (1):
T b + 1 = (a + b)Cb × (p)b
⇒Tb+1=(a+b)!b!(a+b−b)!×pb ⇒Tb+1=(a+b)!b!a!×pb
Therefore, the coefficient of (p)b = (a+b)!b!a! . . . . . . . . . . (3)
Now, on comparing equation (2) and equation (3) we conclude that the coefficients of pa and pb are equal in the expansion of (1 + p)a + b.
Hence, Proved
Q.10: The coefficients of the (k + 1)th, kth and (k – 1)th terms in the expansion of (x + 1)n are in the ratio 5 : 3 : 1. Find the values of n and k.
Sol.
Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Now on substituting a = x and b = 1 in the above equation:
Tr + 1 = nCr × (x)n – r× (1)r
Now, (k + 1)th term in the expansion of (x + 1)n :
T(k) + 1 = nCk × (x)n – k × 1(k )
i.e. T(k + 1) = nCk × (x)n – k
⇒Tk+1=n!k!(n−k)!×(x)n−k
Therefore the coefficient of T(k + 1)th term = n!k!(n−k)!
Now, (k)th term in the expansion of (x + 1)n :
T(k – 1) + 1 = nCk – 1 × (x)n – (k – 1) × 1(k – 1)
i.e. Tk = nCk – 1× (x)n – k + 1
⇒Tk=n!(k−1)!(n−k+1)!×(x)n−k+1
Therefore the coefficient of T(k)th term = n!(k−1)!(n−k+1)!
And, (k – 1)th term in the expansion of (x + 1)n :
T(k – 2 )+ 1 = nCk – 2 × (x)n – (k – 2) × 1(k – 2)
i.e. Tk – 1 = nCk – 2 × (x)n – k + 2
⇒Tk−1=n!(k−2)!(n−k+2)!×(x)n−k+2
Therefore the coefficient of T(k – 1)th term = n!(k−2)!(n−k+2)!
Now, according to the given conditions: Coefficients of T(k + 1), Tk and Tk – 1 are in the ratio of 5 : 3 : 1
Therefore, Tk−1Tk=13
⇒[n!(k−2)!(n−k+2)!]÷[n!(k−1)!(n−k+1)!]=13 ⇒(k−1)!(n−k+1)!(k−2)!(n−k+2)!=13 ⇒(k−1)(k−2)!(n−k+1)!(k−2)!(n−k+2)(n−k+1)!=13 ⇒k−1n−k+2=13 ⇒3k−3=n−k+2
(or) n – 4k + 5 = 0 . . . . . . . . . . . . . . . . (1)
And, Tk+1Tk=53
⇒[n!k!(n−k)!]÷[n!(k−1)!(n−k+1)!]=53 ⇒(k−1)!(n−k+1)!k!(n−k)!=53 ⇒(k−1)![(n−k+1)(n−k)!][k(k−1)!](n−k)!=53 ⇒n−k+1k=53 ⇒3n−3k+3=5k
(or) 3n – 8k + 3 = 0 . . . . . . . . . . . . . . . . . . (2)
On multiplying equation (1) by 3 and subtracting it to equation (2) we will get:
(3n – 8k + 3) – 3(n – 4k + 5) = 0
⇒ 3n – 8k + 3 – 3n + 12k – 15 = 0
⇒ 4k = 12
Therefore, k = 3
Now on substituting the value of k in equation (2) we will get:
3n – 8(3) + 3 = 0
Therefore, n = 7
Q.11: Prove that the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p
Sol.
Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r× (y)r
For (1 + 3a)2p:
On putting n = 2p, x =1 and y = 3a, we will get:
Tr + 1 = (2p)Cr × (1)(3p) – r × (3a)r
⇒ Tr + 1 = (2p)Cr× (3a)r . . . . . . . . . . . . (1)
Now, on comparing the coefficient of a in equation (1) with ap, we will get r = p
Therefore, from equation (1):
T p + 1 = (2p)Cp × (3a)p
⇒Tp+1=(2p)!p!(2p−p)!×(3a)p ⇒Tp+1=(2p)!p!p!×3p×(a)p
Therefore, the coefficient of ap =(2p)!p!p!×3p . . . . . . . . . . . . . . . . (2)
Now, for (1 + 3a)2p – 1 :
On substituting n = (2p – 1), x =1 and y = 3a, we will get:
Tr + 1 = (2p – 1)Cr × (1)(2p – 1) – r × (3a)r
⇒ Tr + 1 = (2p – 1)Cr× (3a)r . . . . . . . . . . . . (1)
Now, on comparing the coefficient of ‘a’ in equation (1) with ‘ap ’, we will get r = p
Therefore, from equation (1):
T p + 1 = (2p – 1)Cp × (3a)p
⇒Tp+1=(2p−1)!p!(2p−1−p)!×(3a)p ⇒Tp+1=(2p−1)!p!(p−1)!×(3)p×(a)p ⇒Tp+1=(2p)!2pp!×p!p×(3)p.(a)p=(2p)!2(p!)2×3p.ap
Therefore, the coefficient of ap = =(2p)!2(p!)2×3p . . . . . . . . . . (3)
Now, on comparing equation (2) and equation (3):
⇒(2p)!p!p!×3p=(2p)!p!p!×3p
⇒ (2p)Cp× 3p = 12 (2p – 1)Cp× 3p
Therefore, the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p
Q.12: For what values of ‘m’ the coefficient of x2 in the expansion of (1+ 2x)k is 140.
Sol.
Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× (b)r
Now, on substituting n = m, a =1 and b = 2x we will get:
Tr + 1 = mCr × (1)m – r × (2x)r
⇒ Tr + 1 = mCr× (2)r× (x)r . . . . . . . . . . . . (1)
Now, on comparing the coefficient of x in equation (1) with x2, we will get r = 2
Therefore, from equation (1):
T 2 + 1 = mC2 × (2)2 × (x)2
Since, the coefficient of x2 = 140 [GIVEN]
Therefore, 140 = mC2 × (2)2
⇒84=[m!2!(m−2)!]×4 ⇒21=m×(m−1)(m−2)![2×1]×(m−2)! ⇒42=m2−m
Therefore, m2 – m – 42 = 0
Now, by splitting of middle term method, the roots of this quadratic equation are:
⇒m2 – (7 – 6)m – 42 = 0
⇒m2 – 7m + 6m – 42 = 0
⇒ m(m – 7) +6(m – 7) = 0
i.e. (m + 6) (m – 7) = 0
Therefore, m = 7 or m = -6
Hence, for the coefficient of x2 in the expansion of (1+ 2x)k to be 140, the value of ‘m’ should be 7.
Q.13: In the expansion of (2x+3y)9, Find the middle terms.
Sol.
Here, n = 9
When n is odd then the middle terms in the expansion of (a + b)n are given by:
middle terms = (n+12)thtermand(n+12+1)thterm
Therefore, the middle terms in the expansion of (2x+3y)9 are:
⇒(9+12)thtermand(9+12+1)thterm
⇒ 5th term and 6th term
Now, 5th and 6th terms in the expansion of (2x+3y)9 are:
T5 = T4 + 1 = 9C4 × (2x)9 – 4 × (3y)4
⇒ T5 = 9C4 × (2x)5 × (3y)4
⇒T5=9!4!5!×(2x)5×(3y)4 ⇒T5=9×8×7×6×5!4×3×2×1×5!×32x5×81y4 ⇒T5=126×2592x5y4=414720x5y4
Therefore, 5th term = 414720 x5 y4
Now, T6 = T5 + 1 = 9C5 × (2x)9 – 5 × (3y)5
\Rightarrow T6 = 9C5 × (2x)4 × (3y)5
\Rightarrow T_{6}=\frac{9!}{5!\times 4!}\times (2x)^{4}\times (3y)^{5} \Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 16x^{4}\times 243y^{5}\\\\ \Rightarrow T_{6} = 126\times 3888\;x^{4}y^{5}\boldsymbol{=489888\;x^{4}y^{5}}\\
Therefore, 6th term = 489888 x4 y5
Hence, 5th and 6th terms are the middle terms in the expansion of (2x+3y)9
And also, 5th term = 414720 x5 y4 and 6th term = 489888 x4 y5
Q.14: Find the Coefficient of x6 in expansion of (x + 3)11
Sol.
Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Therefore, on substituting n = 11, a = x and b = 3 in the above expression we will get:
Tr + 1 = 11Cr × (x)11 – r× 3r . . . . . . . (1)
Now, on comparing the coefficient of x in equation (1) with x7, we will get:
i.e. (x)11 – r = x6
(or) 11 – r = 6
Therefore, r = 5
Now, on substituting r in equation (1) we will get:
T5 + 1 = 11C5 × (x)11 – 5 × 35
\Rightarrow T_{6}=\frac{11!}{5!\times 6!}\times x^{6}\times 3^{5} \Rightarrow T_{6}=\frac{11\times 10\times 9\times 8\times 7\times 6!}{5\times 4\times 3\times 2\times 1\times 6!}\times 243\;x^{6}\\\\ \Rightarrow T_{6} = 462\times 243\;x^{6}\boldsymbol{=112266\;x^{6}}
Therefore, the Coefficient of x6 in the expansion of (x + 3)11 = 112266
Q.15: Write the general term in the expansion of (x2y3 – x3 y2)7
Sol.
The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Now, on substituting n = 7, a = x2y3 and b = x3 y2 we will get:
Tr+1 = 7Cr × (x2y3)7 – r× (x3y2)r
\Rightarrow Tr + 1 = 7Cr × (x2)7 – r × (y3)7 – r × (x)3 r × (y)2 r
\Rightarrow Tr + 1 = 7Cr × (x)14 – 2 r × (x)3 r× (y)2 r× (y)21 – 3r
\Rightarrow Tr + 1 = 7Cr × (x)14 – 2 r + 3r × (y)21 – 3r + 2 r
i.e. Tr + 1 = 7Cr × (x)14 + r × (y)21 – r
Therefore, the general term in the expansion of (x2y3 – x3 y2)7:
Tr + 1 = 7Cr × (x)14 + r× (y)21 – r
Miscellaneous Exercise:
Q.1: If 1st term, 2nd term and 3rd term in the expansion of (a + b)n are 729, 7290 and 30375 respectively then find the value of ‘a’, ‘b’ and ‘c’.
Sol.
The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r× br
Now, according to the given conditions:
T(0 + 1) = nC0 × (a)n – 0 × b0
T1 = an
i.e. an = 729 . . . . . . . . . . . . . . . . . . (1)
Also, T(1 + 1) = nC1 × (a)n – 1 × b1
T2 = nC1 × (a)n – 1 × b
i.e. 7290 = nC1 × (a)n – 1 × b
\\\Rightarrow 7290=\frac{n!}{1!\;\;(n-1)!}\times a^{n}\times a^{-1}\times b\\\\ \\\\\Rightarrow 7290=\frac{n(n-1)!}{(n-1)!}\times a^{n}\times \frac{b}{a}\;\;\;
Therefore, 7290 = n × an× \frac{b}{a} . . . . . . . . . . . . . . . . (2)
And, T(2 + 1) = nC2 × (a)n – 2 × b2
T3 = nC2 × (a)n – 2 × b2
i.e. 30375 = nC2 × (a)n – 2 × b2
\\\Rightarrow 30375=\frac{n!}{2!\;\;(n-2)!}\times a^{n}\times a^{-2}\times b^{2}\\ \\\Rightarrow 30375=\frac{n\;(n-1)\;(n-2)!}{(2\times 1)\;\;(n-2)!}\times a^{n}\times \left ( \frac{b}{a} \right )^{2}\\
Therefore, 60750 = n (n – 1) × an × \left ( \frac{b}{a} \right )^{2} . . . . . . . . . . . . . . . . (3)
Now, on substituting equation (1) in equation (2) we will get:
\\\Rightarrow 7290 = n\times (729)\times \frac{b}{a}\\ \\\Rightarrow \frac{10a}{b}=n\\
Therefore, n = \frac{10a}{b} . . . . . . . . . . . . . . . . . (4)
Now, on dividing equation (2) and equation (3) we will get:
\\\Rightarrow \frac{n \times a^{n} \times \frac{b}{a}}{n\times (n-1)\times a^{n}\times \left ( \frac{b}{a} \right )^{2} }=\frac{7290}{60750}\\ \\\Rightarrow \frac{3}{25}=\frac{1}{(n-1)\times \frac{b}{a}}\\ \\\Rightarrow 3n-3=\frac{25a}{b}\\
Now, from equation (4):
\frac{a}{b} = \frac{n}{10} \Rightarrow 3n-3=25\times \left ( \frac{n}{10} \right ) \Rightarrow 30n-30=25n
i.e. 30n – 30 = 25n
Therefore, n = 6
Now, on substituting the value of ‘n’ in equation (1) we will get:
\Rightarrow a6 = 729
i.e. a6 = 36
Therefore, a = 3
Now, from equation (4):
\\\Rightarrow 6 = \frac{30}{b}\\
Therefore, b = 5
Q.2: If the coefficients of z2 and z3 are equal in the expansion of (3 + bz)9. Find the value of ‘b’
Sol.
Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n –r× yr
Therefore, on substituting n = 9, x = 3 and y = bz in the above expression we will get:
Tr + 1 = 9Cr × (3)9 – r× (bz)r
Tr + 1 = 9Cr × (3)9 – r× br× zr . . . . . (1)
Now, on comparing the coefficient of z in equation (1) with z2, we will get: r = 2
On substituting the value of r in equation (1) we will get:
T(2 + 1) = 9C2 × (3)9 – 2 × b2 × z2
\\\Rightarrow T_{3}=\frac{9!}{2!\;7!}\times 3^{7}\times b^{2}\times z^{2}\\ \\\Rightarrow T_{3}=\frac{9\times 8\times 7!}{2\times 1\times 7!}\times 3^{7}\times b^{2}\times z^{2}\\
\\ \Rightarrow T_{3} = 36\times 3^{7}\times b^{2}\times z^{2} . . . . . . . . . . . . (2)
Now, on comparing the coefficient of z in equation (1) with z3, we will get: r = 3
On substituting the value of ‘r’ in equation (1) we will get:
T(3 + 1) = 9C3 × (3)9 – 3 × b3 × z3
\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times 3^{6}\times b^{3}\times z^{3}\\ \\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 3^{6}\times b^{3}\times z^{3}\\
\\\Rightarrow T_{4} = 84\times 3^{6}\times b^{3}\times z^{3} . . . . . . . . . (3)
Now, according to the given conditions:
Coefficient of z2 = Coefficient of z3
Therefore from equation (2) and equation (3):
\\\Rightarrow 36\times 3^{7}\times b^{2}=84\times 3^{6}\times b^{3}\\ \\\Rightarrow b = \frac{36\times 3^{7}}{84\times 3^{6}}\\
Therefore, the value of b = \frac{9}{7}
Q.3: Find the coefficient of x6 in the product of (1 + 3x)7 (1 – 2x)6 by using binomial theorem.
Sol.
By using Binomial Theorem:
Since, (1 + x)n = [ nC0 ] + [ nC1 × (x) ] + [ nC2 × (x)2 ] + [ nC3 × (x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (x)n ]
Therefore, (1 + 3x)7 = [ 7C0 ] + [ 7C1 × (3x) ] + [ 7C2 × (3x)2 ] + [ 7C3 × (3x)3 ] + [ 7C4 × (3x)4 ] + [ 7C5 × (3x5) ] + [ 7C6 × (3x)6 ] + [ 7C7 × (3x)7 ] \\\Rightarrow 1 + [ 7 × (3x) ] + [ 21 × (9x2) ] + [ 35 × (27x3) ] + [ 35 × (81x4) ] + [ 21 × (243x5) ] + [ 7 × 729x6 ] + [ 1 × 2187x7 ] \\\Rightarrow 1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7
Therefore, (1 + 3x)7 = 1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7
Now, (1 – 2x)6:
By using Binomial Theorem:
(1 – 2x)6 = [ 6C0 ] – [ 6C1 × (2x) ] + [ 6C2 × (2x)2 ] – [ 6C3 × (2x)3 ] + [ 6C4 × (2x)4 ] – [ 6C5 × (2x5) ] + [ 6C6 × (2x)6 ] \\\Rightarrow 1 – [ 6 × (2x) ] + [ 15 × (4x2) ] – [ 20 × (8x3) ] + [ 15 × (16x4) ] – [ 6 × (32x5) ] + [ 1 × 64x6 ] \\\Rightarrow 1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6
Therefore, (1 – 2x)6 = 1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6
Now, (1 + 3x)7 (1 – 2x)6:
=(1 + 21x + 189x2 + 945x3 + 2835x4 + 5103x5 + 5103x6 + 2187x7) × (1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6)
Now, it not required to multiply the above equation, only terms with x5 are required and that can be identified by analyzing the above equation.
\\\Rightarrow [{1 × (- 192x5)} + {(21x) × (240x4)} + {(189x2) × (- 160x3)} + {(945x3) × (60x2)} + {(2835x4) × (- 12x)} + {(5103x5) × (1)}] \\\Rightarrow [ -192x5 + 5040x5 – 30240x5+ 56700x5 – 34020x5 + 5103x5 ]
\\\Rightarrow 2391 x5
Therefore, the coefficient of x5 = 2391
Q.4: Show that (a – b) is a factor of (an – bn), where ‘n’ is a positive integer and ‘a’ and ‘b’ are distinct integers.
Sol.
Now, for (a – b) to be factor of (an – bn), (an – bn) = p(a – b) [where p is any natural number]
We can write an= (a – b + b)n = [(a – b) + b]n
Now, by Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, for x = (a – b) and y = b the equation becomes:
[(a – b) + b]n = [ nC0 × (a – b)n ] + [ nC1 × (a – b)n – 1 × b ] + [ nC2 × (a – b)n – 2 × b2 ] + [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × (a – b) × bn – 1 ] + [ nCn × bn ] \\\Rightarrow [a]n = [(a – b)n] + [ nC1 × (a – b)n – 1 × b ] + [ nC2 × (a – b)n – 2 × b2 ] + [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × (a – b) × bn – 1 ] + [ bn ]
Now, an – bn = [(a – b)n] + [ nC1 × (a – b)n – 1 × b ] + [ nC2 × (a – b)n – 2 × b2 ] + [ nC3 × (a – b)n – 3 × b3 ] + . . . . . . . . . . . . . + [ nCn – 1 × (a – b) × bn – 1 ] + [ bn ] – [ bn ] \\\Rightarrow an – bn = (a – b) × [ (a – b)n – 1 + nC1 (a – b)n – 2 b + nC2 (a – b)n – 3 b2 + nC3 (a – b)n – 4 b3 + . . . . . . . . . . . . . + nCn – 1 bn – 1 ]
\\\Rightarrow (an – bn) = (a – b) × p
Where, p = [ (a – b)n – 1 + nC1 (a – b)n – 2 b + nC2 (a – b)n – 3 b2 + nC3 (a – b)n – 4 b3 + . . . . . . . . . . . . . + nCn – 1 bn – 1 ] is any natural number.
Therefore, (a – b) is a factor of (an – bn), where ‘n’ is a positive integer.
Q.5: Evaluate \left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}
Sol.
Find (a + b)6 – (a – b)6
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, [a + b]6= [ 6C0 × (a6) ] + [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] + [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] + [ 6C5 × a × b5 ] + [6C6 × b6]
And, [a – b]6 = [ 6C0 × (a6) ] – [ 6C1 × (a5) × b ] + [ 6C2 × (a4) × b2 ] – [ 6C3 × (a3) × b3 ] + [ 6C4 × (a2) × b4 ] – [ 6C5 × a × b5 ] + [6C6 × b6]
Therefore, (a + b)6 – (a – b)6 = [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b6] – [ 6C0 a6 – 6C1 a5b + 6C2 a4b2 – 6C3 a3b3 + 6C4 a2b4 – 6C5 ab5 + 6C6 b6 ] \\\Rightarrow [ 6C0 a6 + 6C1 a5b + 6C2 a4b2 + 6C3 a3b3 + 6C4 a2b4 + 6C5 ab5 + 6C6 b6 – 6C0 a6 + 6C1 a5b – 6C2 a4b2 + 6C3 a3b3 – 6C4 a2b4 + 6C5 ab5 – 6C6 b6 ] \\\Rightarrow 2[ 6C1 a5b + 6C3 a3b3 + 6C5 ab5 ] = 2[ 6a5b + 20a3b3 + 6ab5 ]
Therefore, (a + b)6 – (a – b)6 = 4ab[ 3a4 + 10a2 b2 + 3b4 ] . . . . . . (1)
Now, on substituting a = \sqrt{7} and b = \sqrt{5} in equation (1) we will get:
\\\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}:\\
\\=4( \sqrt{7}\times \sqrt{5} )\left [ (3 \sqrt{7} )^{4} +10\times (\sqrt{7} )^{2}\times ( \sqrt{5})^{2}+3(\sqrt{5})^{4} \right ]\\
\\\Rightarrow 4\sqrt{35}\;[\;147+350+75\;]\\
Therefore, \left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}= 2288\sqrt{35}
Q.6: Find the expansion of \left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}
Sol.
The above given expression can be assumed as: (x + y)4 + (x – y)4
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, [x + y]4 = [4C0 × (x4)] + [4C1 × (x3) × (y)] + [4C2 × (x2) × (y)2] + [4C3 × (x1) × (y)3] + [4C4 × (x0) × (y)4]
And, [x – y]4 = [4C0 × (x4) ] – [4C1 × (x3) × (y) ] + [4C2 × (x2) × (y)2 ] – [4C3 × (x1) × (y)3 ] + [4C4 × (x0) × (y)4 ]
Therefore, (x + y)4 + (x – y)4 = [ 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 ] + [ 4C0 x4 – 4C1 x3 y + 4C2 x2 y2 – 4C3 x y3 + 4C4 y4 ] \\\Rightarrow [ 4C0 x4 + 4C1 x3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y4 + 4C0 x4 – 4C1 x3y + 4C2 x2y2 – 4C3 xy3 + 4C4 y4 ] \\\Rightarrow 2[ 4C0 x4 + 4C2 x2 y2 + 4C4 y4] = 2[ x4 + 6x2 y2 + y4 ]
Therefore, (x + y)4 + (x – y)4 = 2[ x4 + y4 + 6 x2 y2 ] . . . . . . . . . . . . . . . . . . . (1)
Now, on substituting x = a3 and y = \sqrt{a^{3}-2} in equation (1) we will get:
\\\Rightarrow \left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}:\\
\\=2\times \left [ (a^{3})^{4}+ (\sqrt{a^{3}-2})^{4}+[6\times (a^{3})^{2}\times (\sqrt{a^{3}-2})^{2}] \right ]\\
\\=2\left [ a^{12}+ (a^{3}-2)^{2}+[6\times a^{6}\times (a^{3}-2)] \right ]\\ \\=2\left [\; a^{12}+ a^{6}+4-4a^{3}+6a^{9}-12a^{6} \;\right ]\\
\\\Rightarrow 2a^{12}+12a^{9}-22a^{6}-8a^{3}+8
Therefore, the expansion of \left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}:
= 2a^{12}+12a^{9}-22a^{6}-8a^{3}+8
Q.7: Find the approximate value of (0.98)5 using the first four terms of its expansion.
Sol.
(0.98)5 = (1 – 0.02)5
Now, by using Binomial Theorem:
Since, [ x – y ]n = [ nC0 × (x)n ] – [ nC1 × (x)n – 1 × y ] + [ nC2 × (x)n – 2 × y2 ] – [ nC3 × (x)n – 3 × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]
Therefore, (1 – 0.02)5 = [ 5C0 × (1)5 ] – [ 5C1 × (1)4 × (0.02) ] + [ 5C2 × (1)3 × (0.02)2 ] – [ 5C3 × (1)2 × (0.02)3 ]
= 1 – [5 × 1 × 0.02] + [5 × 1 × 0.0004] – [5 × 1 × 0.000008]
= 1 – 0.10 + 0.0020 + 0.000040
= 0.90204
Therefore, (0.98)5 = 0.90204
Q.8: The coefficient of the 5th term from end and the 5th term from the beginning in the expansion of \left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n} are in the ratio 1:\sqrt{6}.Find the value of ‘n’.
Sol.
Since, The general term in the expansion of (a + b)n is given by: Tk + 1 = nCk × (a)n – k× bk
Now on substituting a = \left ( 2\right )^{{\frac{1}{4}}} and b = \frac{1}{(3)^{\frac{1}{4}}} in the above equation:
Tk + 1 = nCk ×\left ( \sqrt[4]{2} \right )^{n-k} \times \left ( \frac{1}{\sqrt[4]{3}}\right )^{k}\\
Now, 5th term from beginning in the expansion of \left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}:
T(4 + 1) = nC4 × \left ( \sqrt[4]{2} \right )^{n-4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}\\
i.e. T5 = nC4 × \left ( 2 \right )^{\frac{n}{4}-1} \times \left ( \frac{1}{3} \right )\\
\\\Rightarrow T_{5} = \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times (2)^{-1}\times \frac{1}{3}\\
Therefore the coefficient of 5th term from beginning \boldsymbol{= \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6}}\\
Now, 5th term from end in the expansion of \left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}:
1st term, 2nd term, 3rd term . . . . . . . . . . . (n – 5)th term, (n – 4)th term, (n – 3)th term, (n – 2)th term, (n – 1)th term, nth term.
Therefore, 5th term from end will be (n – 4)th term from beginning.
i.e. T(n – 4) + 1 = nCn – 4 × \left ( \sqrt[4]{2} \right )^{n-(n-4)} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}\\
i.e. T(n – 3) = nCn – 4× \left ( \sqrt[4]{2} \right )^{4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n}\times \left ( \frac{1}{\sqrt[4]{3}} \right )^{-4}\\
\\\Rightarrow T_{(n-3)} = \frac{n!}{(n-4)!\;[(n-(n-4)]!}\times 2 \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 3\\
Therefore the coefficient of 5th term from end [ T(n – 4)th term ] = \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6\\
Now, according to the given conditions:
The ratio of coefficient of 5th term from end and the coefficient of 5th term from beginning is 1:\sqrt{6}\\
Therefore, \\ \frac{T_{n-3}}{T_{5}}=\frac{1}{\sqrt{6}}\\
\\\Rightarrow \left ( \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6 \right )\div\left ( \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6} \right )=\frac{1}{\sqrt{6}}\\ \\\Rightarrow \frac{4!\;(n-4)!}{(n-4)!\;4!}\times \frac{1}{(2)^{\frac{n}{4}}}\times \frac{1}{(3)^{\frac{n}{4}}}\times 6\times 6=\frac{1}{\sqrt{6}}\\ \\\Rightarrow \frac{1}{(6)^{\frac{n}{4}}}\times 36=\frac{1}{\sqrt{6}}\\ \\\Rightarrow 36\sqrt{6}=(6)^{\frac{n}{4}}\\ \\\Rightarrow (6)^{2}\times (6)^{\frac{1}{2}}=6^{\frac{n}{4}}\\ \\\Rightarrow (6)^{\frac{5}{2}}=(6)^{\frac{n}{4}\\}
On comparing RHS and LHS:
\\\Rightarrow \frac{5}{2}=\frac{n}{4}\\
\\\Rightarrow n = 10
Therefore, the value of n = 10.
Q.9: By using the Binomial Theorem, find the expansion of \left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}; where ≠ 0
Sol.
The given expression \left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4} can be expanded as \left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, \left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4} = [4C0 × \left (1-\frac{3}{x} \right )^{4}] + [4C1 × \left (1-\frac{3}{x} \right )^{3} × \left ( \frac{x}{3} \right )] + [4C2 × \left (1-\frac{3}{x} \right )^{2} × \left ( \frac{x}{3} \right )^{2}] + [4C3 × \left (1-\frac{3}{x} \right )^{1} × \left ( \frac{x}{3} \right )^{3} + [4C4 × \left ( \frac{x}{3} \right )^{4}]
\\=\left [ 1\times \left (1-\frac{3}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{3}\times \left ( \frac{x}{3} \right ) \right ]+\left [ 6\times \left (1-\frac{3}{x} \right )^{2} \times \left ( \frac{x}{3} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{1} \times \left ( \frac{x}{3} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{3} \right )^{4} \right ] . . . . . . . . . . . . . . . . . . . (1)
Now, \left (1-\frac{3}{x} \right )^{4}:
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]
Therefore, \left (1-\frac{3}{x} \right )^{4} = [4C0] – [4C1 × \left ( \frac{3}{x} \right )^{1}] + [4C2 × \left ( \frac{3}{x} \right )^{2}] – [4C3 × \left ( \frac{3}{x} \right )^{3}] + [4C4 × \left ( \frac{3}{x} \right )^{4}] \\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{3}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{27}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{81}{x^{4}} \right ) \right ]\\
\\=\left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}} \right ] . . . . . . . . (2)
And, \left (1-\frac{3}{x} \right )^{3} = [ 3C0 ] – [ 3C1 × \left ( \frac{3}{x} \right )^{1}] + [3C2 × \left ( \frac{3}{x} \right )^{2}] – [3C3 × \left ( \frac{3}{x} \right )^{3}] \\= [1]-\left [ 3\times \left ( \frac{3}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{27}{x^{3}} \right ) \right ]\\ \\=\left [1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right ] . . . . . . . . . . . . (3)
Now, from equation (1), equation (2) and equation (3):
\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ 4 \times \left ( 1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right )\times \frac{x}{3} \right ]+\left [ 6\times \left (1+\frac{9}{x^{2}}-\frac{6}{x} \right )\times \frac{x^{2}}{9}\right ] +\left [ \left ( 4-\frac{12}{x} \right )\times \frac{x^{3}}{27} \right ]+\left [\frac{x^{4}}{81}\right ]\\ \\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ \frac{4x}{3}-12+\frac{36}{x}-\frac{36}{x^{2}}\right ]+\\\\\\\\+\left [ \frac{2x^{2}}{3}+6-4x\right ] +\left [ \frac{4x^{3}}{27}-\frac{4x^{2}}{9} \right ]+\left [\frac{x^{4}}{81}\right ]\\ \\\boldsymbol{\Rightarrow \left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\\
Therefore, the expansion of \left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}:
= \boldsymbol{\left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}
Q.10: By using the Binomial Theorem, find the expansion of (6x2 – 4ax + 6a2)3.
Sol.
The given expression (6x2 – 4ax + 6a2)3 can be expanded as [(6x2 – 4ax) + 6a2)3]
By using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, [(6x2 – 4ax) + 6a2)3] = [3C0 × (6x2 – 4ax)3] + [3C1 × (6x2 – 4ax)2 × 6a2] + [3C2 × (6x2 – 4ax)1 × (6a2)2] + [3C3 × (6x2 – 4ax)0 × (6a2)3]
= [1× (6x2 – 4ax)3] + [3 × (6x2 – 4ax)2 × 6a2] + [3 × (6x2 – 4ax) × (6a2)2] + [1 × (6a2)3]
= [(6x2 – 4ax)3] + [(6x2 – 4ax)2 × 18a2] + [(6x2 – 4ax) × 108a4 ] + [216a6] . . . . . . . . . . . . (1)
Now, [(6x2 – 4ax)3]:
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]
Therefore, [(6x2 – 4ax)3] = [3C0 × (6x2)3] – [3C1 × (6x2)2) × 4ax] + [3C2 × 6x2 × (4ax)2] – [3C3 × (4ax)3]
= [1 × (216x6)] – [3 × (36x4) × 4ax] + [3 × (6x2)× 16a2x2] – [1 × (64a3x3)]
\Rightarrow [(6x2 – 4ax)3] = [216x6 – 432 ax5 + 288 a2x4 – 64a3x3] . . . . . . . . . . (2)
From equation (1) and equation (2)
=[216x6 – 432 ax5 + 288 a2x4 – 64a3x3] + [(36x4 + 16a2x2 – 48ax3) × 18a2] + [(6x2 – 4ax) × 108a4 ] + [ 216a6 ]
=[216x6 – 432 ax5 + 288 a2x4 – 64a3x3 + 648a2x4 + 288a4x2 – 864a3x3 + 648x2a4 – 432xa5 + 216a6]
=8[27x6 – 54 ax5 + 36 a2x4 – 8a3x3 + 81a2x4 + 36a4x2 – 108a3x3 + 81x2a4 – 54xa5 + 27a6]
=8[27x6 – 54 ax5 + 117a2 x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6]
Therefore, the expansion of (6x2 – 4ax + 6a2)3 :
= 8[27x6 – 54 ax5 + 117a2 x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6]
Q.11: By using the Binomial Theorem, find the expansion of \left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}; where ≠ 0
Sol.
The given expression \left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4} can be expanded as \left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, \left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}
= [4C0 × \left (1-\frac{5}{x} \right )^{4}] + [4C1 × \left (1-\frac{5}{x} \right )^{3} × \left ( \frac{x}{5} \right )] + [4C2 × \left (1-\frac{5}{x} \right )^{2} × \left ( \frac{x}{5} \right )^{2}] + [4C3 × \left (1-\frac{5}{x} \right )^{1} × \left ( \frac{x}{5} \right )^{3} + [4C4 × \left ( \frac{x}{5} \right )^{4}]
\\=\left [ 1\times \left (1-\frac{5}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{3}\times \left ( \frac{x}{5} \right ) \right ]+\left [ 6\times \left (1-\frac{5}{x} \right )^{2} \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{1} \times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{5} \right )^{4} \right ] . . . . . . . . . . . . . . . . . . . (1)
Now, \left (1-\frac{5}{x} \right )^{4}:\\
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]
Therefore, \left (1-\frac{5}{x} \right )^{4} = [4C0] – [4C1 × \left ( \frac{5}{x} \right )^{1}] + [4C2 × \left ( \frac{5}{x} \right )^{2}] – [4C3 × \left ( \frac{5}{x} \right )^{3}] + [4C4 × \left ( \frac{5}{x} \right )^{4}] \\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{5}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{125}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{625}{x^{4}} \right ) \right ]\\
\\=\left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ] . . . . . . . . (2)
And, \left (1-\frac{5}{x} \right )^{3}:\\
[ 3C0 ] – [ 3C1 × \left ( \frac{5}{x} \right )^{1}] + [3C2 × \left ( \frac{5}{x} \right )^{2}] – [3C3 × \left ( \frac{5}{x} \right )^{3}] \\= [1]-\left [ 3\times \left ( \frac{5}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{125}{x^{3}} \right ) \right ]\\
\\=\left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ] . . . . . . . . . . . . . . . . . . . (3)
Now, from equation (1), equation (2) and equation (3):
\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ 4 \times \left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ] \times \left ( \frac{x}{5} \right ) \right ]+\\\\\\+\left [ 6\times \left (1+\frac{25}{x^{2}} -\frac{10}{x}\right ) \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )\times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ \left ( \frac{x}{5} \right )^{4} \right ]\\ \\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ \frac{4x}{5}-12+\frac{60}{x}-\frac{100}{x^{2}} \right ]+\left [ \frac{6x^{2}}{25}+6-\frac{12x}{5}\right ]+\left [ \frac{4x^{3}}{125}-\frac{x^{2}}{25} \right ]+\left [ \frac{x^{4}}{625} \right ]\\ \\\\\Rightarrow \left [\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}-\frac{8x}{5}-5+\frac{40}{x}+ \frac{x^{2}}{5}+\frac{4x^{3}}{125}+\frac{x^{4}}{625} \right ]\\\\
Therefore, the expansion of \left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}:
= \left [\frac{x^{4}}{625}+\frac{4x^{3}}{125}+ \frac{x^{2}}{5}-\frac{8x}{5}-5+\frac{40}{x}+\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}\right ]
Q.12: By using the Binomial Theorem, find the expansion of (5x3 – 2ax + 3a3)3.
Sol.
The given expression (5x3 – 2ax + 3a3)3 can be expanded as [(5x3 – 2ax) + 3a3)3]
By using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
Therefore, [ (5x3 – 2ax) + 3a3)3 ] = [ 3C0 × (5x3 – 2ax)3 ] + [ 3C1 × (5x3 – 2ax)2 × 3a3 ] + [ 3C2 × (5x3 – 2ax)1 × (3a3)2 ] + [ 3C3 × (5x3 – 2ax)0 × (3a3)3 ]
= [1× (5x3 – 2ax)3] + [3 × (5x3 – 2ax)2 × 3a3] + [3 × (5x3 – 2ax) × (3a3)2] + [1 × (3a3)3]
= [(5x3 – 2ax)3] + [(5x3 – 2ax)2 × 9a3] + [(5x3 – 2ax) × 27a6] + [27a9] . . . . . . . . . . . . (1)
Now, [(5x3 – 2ax)3]:
By using Binomial Theorem:
Since, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . (-1)n [ nCn × yn ]
Therefore, [(5x2 – 2ax)3] = [3C0 × (5x3)3] – [3C1 × (5x3)2) × 2ax] + [3C2× 5x3 × (2ax)2] – [3C3× (2ax)3]
= [1 × 125x9] – [3 × (25x6)× 2ax] + [3 × (5x3) × 4a2x2] – [1 × 8a3x3]
[(5x3 – 2ax)3] = [125x6 – 150ax5 + 60a2x4 – 8a3x3] . . . . . . . . . . (2)
From equation (1) and equation (2):
=[125x6 – 150ax5 + 60a2x4 – 8a3x3] + [(25x6 + 4a2x2 – 20ax4) × 9a3] + [(5x3 – 2ax) × 27a6 ] + [ 27a9 ]
=[125x6 – 150ax5 + 60a2x4 – 8a3x3] + [225a3x6 + 36a5x2 – 180a4x4] + [135a6x3 – 54a7x ] + [ 27a9 ]
=[125x6 – 150ax5 + 60a2x4 – 8a3x3 + 225a3x6 + 36a5x2 – 180a4x4 + 135a6x3 – 54a7x + 27a9 ]
=[125x6 + 225a3x6 – 150ax5 + 60a2x4 – 180a4x4 – 8a3x3 + 135a6x3 + 36a5x2 – 54a7x + 27a9 ]
Therefore, the expansion of (5x3 – 2ax + 3a3)3 :
=[125x6 + 225a3x6 – 150ax5 + 60a2x4 – 180a4x4 – 8a3x3 + 135a6x3 + 36a5x2 – 54a7x + 27a9]
Q.13: Find the coefficient of x7 in the product of (1 + x)7 (1 + 3x)6 by using binomial theorem.
Sol.
By using Binomial Theorem:
Since, (1 + x)n = [ nC0 ] + [ nC1 × (x) ] + [ nC2 × (x)2 ] + [ nC3 × (x3) ] + . . . . . . . . . . . . . . . . . . + [ nCn × (x)n ]
Therefore, (1 + x)7 = [ 7C0 ] + [ 7C1 × (x) ] + [ 7C2 × (x)2 ] + [ 7C3 × (x)3 ] + [ 7C4 × (x)4 ] + [ 7C5 × (x5) ] + [ 7C6 × (x)6 ] + [ 7C7 × (x)7
\\\Rightarrow 1 + [ 7x ] + [ 21x2 ] + [ 35x3 ] + [ 35x4 ] + [ 21x5 ] + [ 7x6 ] + [ x7 ]
Therefore, (1 + x)7 = 1+7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7
Now, (1 + 3x)6:
By using Binomial Theorem:
(1 + 3x)6 = [ 6C0 ] + [ 6C1 × (3x) ] + [ 6C2 × (3x)2 ] + [ 6C3 × (3x)3 ] + [ 6C4 × (3x)4 ] + [ 6C5 × (3x5) ] + [ 6C6 × (3x)6 ] \\\Rightarrow 1 + [ 6 × (3x) ] + [ 15 × (9x2) ] + [ 20 × (27x3) ] + [ 15 × (81x4) ] + [ 6 × (243x5) ] + [ 1 × 729x6 ] \\\Rightarrow 1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6
Therefore, (1 + 3x)6 = 1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6
Now, (1 + x)7 (1 + 3x)6:
(1+7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7) × (1 + 18x + 135x2 + 540x3 + 1215x4 + 1458x5 + 729x6)
Now, it not required to multiply the above equation, only terms with x5 are required and that can be identified by analysing the above equation.
\\\Rightarrow [{(7x) × (729x6)} + {(21x2) × (1458x5)} + {(35x3) × (1215x4)} + {(35x4) × (540x3)} + {(21x5) × (135x2)} + {(7x6) × (18x)} + {(x7) × (1)}] \\\Rightarrow [5103x5 + 30618x5 + 42525x5+ 18900x5 + 2835x7 + 126x7 + x7]
\\\Rightarrow\\ 72551 x7
Therefore, the coefficient of x7 = 72551
Q.14: Evaluate \left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}
Sol.
Find (a + b)5 – (a – b)5
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, [a + b]5 = [ 5C0 × (a5) ] + [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] + [ 5C3 × (a2) × b3 ] + [ 5C4 × (a1) × b4 ] + [ 5C5 × b5 ]
And, [a – b]5 = [ 5C0 × (a5) ] – [ 5C1 × (a4) × b ] + [ 5C2 × (a3) × b2 ] – [ 5C3 × (a2) × b3 ] + [ 5C4 × (a) × b4 ] – [ 5C5 × b5 ]
Therefore, (a + b)5 – (a – b)5 = [5C0 a5 + 5C1 a4b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5] – [5C0 a5 – 5C1 a4 b + 5C2 a3b2 – 5C3 a2b3 + 5C4 ab4 – 5C5 ab5] \\\Rightarrow [5C0 a5 + 5C1 a4b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5 – 5C0 a5 + 5C1 a4 b – 5C2 a3b2 + 5C3 a2b3 – 5C4 ab4 + 5C5 ab5] \\\Rightarrow 2[ 5C1 a4b + 5C3 a2b3 + 5C5 b5 ] = 2[ 5a4b + 10a2b3 + b5 ]
Therefore, (a + b)5 – (a – b)5 = 2b[ 5a4 + 10a2 b2 + b4 ] . . . . . . (1)
Now, on substituting a = \sqrt{3} and b = \sqrt{7} in equation (1) we will get:
\\\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}:\\
\\=2(\sqrt{7})\times \left [ 5( \sqrt{3})^{4} +10\times ( \sqrt{3})^{2}\times (\sqrt{7})^{2}+(\sqrt{7})^{4} \right ]\\
\\\Rightarrow 2\sqrt{7}\left [\;45+210+49\; \right ]\\
Therefore, \left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}= 608\sqrt{7}
Q.15: Find the expansion of \left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}
Sol.
The above given expression can be assumed as: (x + y)4 + (x – y)4
Now, by using Binomial Theorem:
Since, [x + y]n = [ nC0 × (xn) ] + [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] + [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + [ nCn × yn ]
And, [x – y]n = [ nC0 × (xn) ] – [ nC1 × (xn – 1) × y ] + [ nC2 × (xn – 2) × y2 ] – [ nC3 × (xn – 3) × y3 ] + . . . . . . . . . . . . . . . . . . + (-1)n [ nCn × yn ]
Therefore, [x + y]4 = [4C0 × (x4)] + [4C1 × (x3) × (y)] + [4C2 × (x2) × (y)2] + [4C3 × (x1) × (y)3] + [4C4 × (x0) × (y)4]
And, [x – y]4 = [4C0 × (x4) ] – [4C1 × (x3) × (y) ] + [4C2 × (x2) × (y)2 ] – [4C3 × (x1) × (y)3 ] + [4C4 × (x0) × (y)4 ]
Therefore, (x + y)4 + (x – y)4 = [ 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 ] + [ 4C0 x4 – 4C1 x3 y + 4C2 x2 y2 – 4C3 x y3 + 4C4 y4 ] \\\Rightarrow [ 4C0 x4 + 4C1 x3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y4 + 4C0 x4 – 4C1 x3y + 4C2 x2y2 – 4C3 xy3 + 4C4 y4 ] \\\Rightarrow 2[ 4C0 x4 + 4C2 x2 y2 + 4C4 y4] = 2[ x4 + 6x2 y2 + y4 ]
Therefore, (x + y)4 + (x – y)4 = 2[ x4 + y4 + 6 x2 y2 ] . . . . . . . . . . . . . . . . . . . (1)
Now, on substituting x = a2 and y = \sqrt{a^{2}-5} in equation (1) we will get:
\\\Rightarrow \left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}:\\
\\=2\times \left [ (a^{2})^{4}+ \left ( \sqrt{a^{2}-5} \right )^{4}+6\times (a^{2})^{2}\times \left ( \sqrt{a^{2}-5} \right )^{2} \right ]\\ \\\Rightarrow 2\left [ a^{8}+ (a^{2}-5)^{2}+[6a^{4}\times (a^{2}-5)] \right ]\\ \\\Rightarrow 2 \left [ a^{8}+a^{4}+ 25 + 10a^{2}+6a^{6}-30a^{4} \right ]\\
Therefore, the expansion of \left [ a^{2} +\sqrt{a^{2}-5}\right ]^{4}+ \left [ a^{2} -\sqrt{a^{2}-5}\right ]^{4}:
=2 \left [ a^{8}+6a^{6}-29a^{4}+10a^{2}+25 \right ]
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