Class 11 Maths Ncert Solutions Chapter 8 Ex 8.3 Binomial Theorem PDF

Class 11 Maths Ncert Solutions Ex 8.3

Class 11 Maths Ncert Solutions Chapter 8 Ex 8.3

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

7290=n!1!(n1)!×an×a1×b 7290=n(n1)!(n1)!×an×ba

Therefore, 7290 = n × an × ba . . . . . . . . . . . . . . . . (2)

And, T(2 + 1) = nC2 × (a)n – 2 × b2

T3 = nC2 × (a)n – 2 × b2

i.e.   30375 = nC2 × (a)n – 2 × b2

30375=n!2!(n2)!×an×a2×b2 30375=n(n1)(n2)!(2×1)(n2)!×an×(ba)2

Therefore, 60750 = n (n – 1) × an × (ba)2  . . . . . . . . . . . . . . . . (3)

Now, on substituting equation (1) in equation (2) we will get:

7290=n×(729)×ba 10ab=n

Therefore, n = 10ab . . . . . . . . . . . . . . . . . (4)

Now, on dividing equation (2) and equation (3) we will get:

n×an×ban×(n1)×an×(ba)2=729060750 325=1(n1)×ba 3n3=25ab

Now, from equation (4):

ab = n10

3n3=25×(n10) 30n30=25n

i.e.   30n – 30 = 25n

Therefore, n = 6

Now, on substituting the value of ‘n’ in equation (1) we will get:

a6 = 729

i.e.  a6 = 36

Therefore, a = 3

Now, from equation (4):

6 = 30b

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

T3=9!2!7!×37×b2×z2 T3=9×8×7!2×1×7!×37×b2×z2

T3=36×37×b2×z2 . . . . . . . . . . . . (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

T4=9!3!6!×36×b3×z3 T4=9×8×7×6!3×2×1×6!×36×b3×z3

T4=84×36×b3×z3.........(3)

Now, according to the given conditions:

Coefficient of z2 = Coefficient of z3

Therefore from equation (2) and equation (3):

36×37×b2=84×36×b3 b=36×3784×36

Therefore, the value of b = 97

 

 

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)]  +  [ 7C3 × (3x)]  + [ 7C4 × (3x)4 ]  +  [ 7C5 × (3x5) ]  +  [ 7C6 × (3x)6 ]  +  [ 7C7 × (3x)7 ]

1 + [ 7 × (3x) ] + [ 21 × (9x2) ] + [ 35 × (27x3) ] + [ 35 × (81x4) ] + [ 21 × (243x5) ] + [ 7 × 729x6 ] + [ 1 × 2187x7 ]

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)]  –  [ 6C3 × (2x)]  + [ 6C4 × (2x)4 ]  –  [ 6C5 × (2x5) ]  +  [ 6C6 × (2x)6 ]

1 – [ 6 × (2x) ] + [ 15 × (4x2) ] – [ 20 × (8x3) ] + [ 15 × (16x4) ] – [ 6 × (32x5) ] + [ 1 × 64x6 ]

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.

[{1 × (- 192x5)} + {(21x) × (240x4)} + {(189x2) × (- 160x3)} + {(945x3) × (60x2)} + {(2835x4) × (- 12x)} + {(5103x5) × (1)}]

[ -192x5 + 5040x5 – 30240x5+ 56700x5 – 34020x5 + 5103x5 ]

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 ]

[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 ]

   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 ]

(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 (7+5)6(75)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 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 = 7 and b = 5 in equation (1) we will get:

(7+5)6(75)6:

=4(7×5)[(37)4+10×(7)2×(5)2+3(5)4]

435[147+350+75]

Therefore, (7+5)6(75)6= 228835

 

 

Q.6: Find the expansion of (a3+a32)4+(a3a32)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 x44C1 x3 y + 4C2 x2 y24C3 x y3 + 4C4 y4 ]

[ 4C0 x4 + 4C1 x3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y4 + 4C0 x44C1 x3y + 4C2 x2y24C3 xy3 + 4C4 y4 ]

  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 = a32 in equation (1) we will get:

(a3+a32)4+(a3a32)4:

=2×[(a3)4+(a32)4+[6×(a3)2×(a32)2]]

 =2[a12+(a32)2+[6×a6×(a32)]]

=2[a12+a6+44a3+6a912a6]

2a12+12a922a68a3+8

Therefore, the expansion of (a3+a32)4+(a3a32)4:

= 2a12+12a922a68a3+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  (214+1314)n are in the ratio 1:6.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 = (2)14 and b = 1(3)14 in the above equation:

Tk + 1 = nCk ×(24)nk×(134)k

Now, 5th term from beginning in the expansion of (214+1314)n:

T(4 + 1) = nC4 × (24)n4×(134)4

i.e.  T5 = nC4 × (2)n41×(13)

T5=n!4!(n4)!×(2)n4×(2)1×13

Therefore the coefficient of 5th term from beginning =n!4!(n4)!×(2)n4×16

Now, 5th term from end in the expansion of (214+1314)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 × (24)n(n4)×(134)n4

i.e.  T(n – 3) = nCn – 4 × (24)4×(134)n×(134)4

T(n3)=n!(n4)![(n(n4)]!×2×(13)n4×3

Therefore the coefficient of 5th term from end [ T(n – 4)th term ] =n!(n4)![(4)]!×(13)n4×6

Now, according to the given conditions:

The ratio of coefficient of 5th term from end and the coefficient of 5th term from beginning is 1:6

Therefore, Tn3T5=16

(n!(n4)![(4)]!×(13)n4×6)÷(n!4!(n4)!×(2)n4×16)=16 4!(n4)!(n4)!4!×1(2)n4×1(3)n4×6×6=16 1(6)n4×36=16 366=(6)n4 (6)2×(6)12=6n4 (6)52=(6)n4

On comparing RHS and LHS:

52=n4

n=10

Therefore, the value of n = 10.

 

 

Q.9: By using the Binomial Theorem, find the expansion of (1+x33x)4;     where 0

 

Sol.

The given expression (1+x33x)4 can be expanded as [(13x)+x3]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, [(13x)+x3]4 = [4C0 × (13x)4]  +  [4C1 × (13x)3 × (x3)]  +  [4C2 × (13x)2 × (x3)2]  +  [4C3 × (13x)1 × (x3)3  +  [4C4 × (x3)4]

=[1×(13x)4]+[4×(13x)3×(x3)]+[6×(13x)2×(x3)2]+[4×(13x)1×(x3)3]+[1×(x3)4] . . . . . . . . . . . . . . . . . . . (1)

Now, (13x)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, (13x)4 = [4C0]  –  [4C1 × (3x)1]  +  [4C2 × (3x)2]  –  [4C3 × (3x)3]  +  [4C4 × (3x)4]

=[1][4×(3x)]+[6×(9x2)][4×(27x3)]+[1×(81x4)]

=[112x+54x2108x3+81x4] . . . . . . . . (2)

And, (13x)3 = [ 3C0 ]  –  [ 3C1 × (3x)1]  +  [3C2 × (3x)2]  –  [3C3 × (3x)3]

=[1][3×(3x)]+[3×(9x2)][1×(27x3)]

=[19x+27x227x3] . . . . . . . . . . . . (3)

Now, from equation (1), equation (2) and equation (3):

[112x+54x2108x3+81x4]+[4×(19x+27x227x3)×x3]+[6×(1+9x26x)×x29]+[(412x)×x327]+[x481] [112x+54x2108x3+81x4]+[4x312+36x36x2]++[2x23+64x]+[4x3274x29]+[x481] [x481+4x327+2x298x35+24x+18x2108x3+81x4]

Therefore, the expansion of (1+x33x)4:

= [x481+4x327+2x298x35+24x+18x2108x3+81x4]

 

 

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)]

[(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 (1+x55x)4; where 0

 

Sol.

The given expression (1+x55x)4 can be expanded as [(15x)+x5]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, [(15x)+x5]4

= [4C0 × (15x)4]  +  [4C1 × (15x)3 × (x5)]  +  [4C2 × (15x)2 × (x5)2]  +  [4C3 × (15x)1 × (x5)3  +  [4C4 × (x5)4]

=[1×(15x)4]+[4×(15x)3×(x5)]+[6×(15x)2×(x5)2]+[4×(15x)1×(x5)3]+[1×(x5)4] . . . . . . . . . . . . . . . . . . . (1)

Now, (15x)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, (15x)4 = [4C0]  –  [4C1 × (5x)1]  +  [4C2 × (5x)2]  –  [4C3 × (5x)3]  +  [4C4 × (5x)4]

=[1][4×(5x)]+[6×(25x2)][4×(125x3)]+[1×(625x4)]

=[120x+150x2500x3+625x4] . . . . . . . . (2)

And, (15x)3:

[ 3C0 ]  –  [ 3C1 × (5x)1]  +  [3C2 × (5x)2]  –  [3C3 × (5x)3]

=[1][3×(5x)]+[3×(25x2)][1×(125x3)]

=[115x+75x2125x3] . . . . . . . . . . . . . . . . . . . (3)

Now, from equation (1), equation (2) and equation (3):

[120x+150x2500x3+625x4]+[4×[115x+75x2125x3]×(x5)]++[6×(1+25x210x)×(x5)2]+[4×(15x)×(x5)3]+[(x5)4] [120x+150x2500x3+625x4]+[4x512+60x100x2]+[6x225+612x5]+[4x3125x225]+[x4625] [50x2500x3+625x48x55+40x+x25+4x3125+x4625]

Therefore, the expansion of (1+x55x)4:

= [x4625+4x3125+x258x55+40x+50x2500x3+625x4]

 

 

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