Class 11 Maths Ncert Solutions Ex 8.1

Class 11 Maths Ncert Solutions Chapter 8 Ex 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 (5xx5)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, (5xx5)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, (5xx5)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 (1xx2)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, (1xx2)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, (1xx2)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(53)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 = [ 5Ca5  +  5C1 a4 b  +  5C2 a3b2  +  5C3 a2b3  +  5C4 ab4  +   5C5 b5 ]  –  [ 5C0 a5  –  5C1 a4b  +  5C2 a3b2  –  5C3 a2b3  5C4 ab4  –  5C5 b5 ]

 [ 5Ca5  +  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(53)5=23×[5(5)4+10(5)2×(3)2+(3)4]

23[125+150+9]=5683

Therefore, (5+3)5(53)5=5683

 

 

Q.12:  Find (1 + x)5 – (1 – x)5 . Hence evaluate (1+7)5(17)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(17)5=27[5+10(7)2+(7)4] 27[5+70+49]=2487

Therefore, (1+7)5(17)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(23)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 a4b6C3 a3b3  +  6C4 a2b4  +  6C5 ab5  +  6C6 b6] – [ 6C0 a6  – 6C1 a5b  +  6C2 a4b2  –  6C3 a3b6C4 a2b4  –  6C5 ab5  +  6C6 b6 ]

 [ 6C0 a6  +  6C1 a5b  +  6C2 a4b6C3 a3b3  +  6C4 a2b4  +  6C5 ab5  +  6C6 b6  –  6C0 a6  + 6C1 a5b  –  6C2 a4b2  +  6C3 a3b–  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(23)6=4(2.3)[3(2)4+10(2)2×(3)2+3(3)4] 46[12+60+27]=3966

Therefore, (2+3)6(23)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 (32xx5)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, (32xx5)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, (32xx5)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)= [ 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