# Binomial Theorem – Exercise 8.2 – Class XI

1. Find the coefficient of x5 in (x + 3)8

Solution:

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-r br

Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain

Tr+1  = 8Cr (x)8-r(3)r

Comparing the indices of x in x5 and in Tr +1,

we obtain r = 3

Thus, the coefficient of x5 is 8C3(3)3 = 8!/3!5! x 33 = 8.7.6.5!/3.2.5! .33 = 1512

1. Find the coefficient of a5b7 in (a – 2b)12

Solution:

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-r br

Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, obtain

Tr+1 = 12Cr(a)12-r(-2b)r = 12Cr (-2)r(a)12-r(b)r

Comparing the indices of a and b in a5b7 and in Tr+1

we obtain r = 7

Thus the coefficient of a5b7 is

12C7(-2)7 = –12!/7!5! .27 = –12.11.10.9.8.7!/5.4.3.2.7! .27 = -(792)(128) = -101376

1. Write the general term in the expansion of (x2 – y)6

Solution:

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-r br

Thus the general term in the expansion of (x2 – y)6 is

Tr+1 = 6Cr(x2)6-r(-y)r = (-1)r 6Cr (x)12-2r(y)r

1. Write the general term in the expansion of (x2 – yx)12, x≠0

Solution:

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-r br

Thus the general term in the expansion of (x2 – yx)12is

Tr+1 = 12Cr(x2)12-r(-yx)r = (-1)r 12Cr (x)24-r(y)r

1. Find the 4th term in the expansion of (x – 2y)12

Solution:

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by Tr+1 = nCr an-r br

Thus, the 4th term in the expansion of (x – 2y)12

T4 = T3+1 = 12C3 (x)12-3(-2y)3 = (-1)3 12!/3!9! .x9.(2)3.y3 = – 12.11.10/3.2 .(2)3.x9y3 = -1760x9y3

1. Find the 13th term in the expansion of (9x – 1/3√x)18, x≠0

Solution:

It is known that (r+1)th term, (Tr+1) in the binomial expansion of (a+b)n is given by Tr+1 = nCr an-r br

The 13th term in the expansion of (9x – 1/3√x)18, x≠0

T13 = T12+1 = 18C12 (9x)18-12 (– 1/3√x)12

=(-1)12 . 18!/12!6! (9)6(x)6 (1/3)12(1/√x)12

= 18.17.16.15.14.13.12!/12!6.5.4.3.2. x6.(1/x6).312.(1/312)

= 18564

1. Find the middle terms in the expansion of [3 – (x3/6)]7

Solution:

It is known that in the expansion of (a+b)n, if n is odd, then there are two middle terms, namely (n+1/2)th term and (n+1/2 + 1)th term.

Therefore the middle terms in the expansion [3 – (x3/6)]7 are (7+1/2)th = 4th term and (7+1/2 + 1)th = 5th term

T4 = T3+1 = 7C3 (3)7-3 [-(x3/6)]3 = (-1)3 7!/3!4! . 34.(x9/63)

= –7.6.5.4!/3.2.4! . 34.[1/23.33].x9 = –105/8 .x9

T5 = T4+1 = 7C4 (3)7-4 [-(x3/6)]4 = (-1)4 7!/3!4! . 33.(x12/64)

= –7.6.5.4!/3.2.4! . [(33)/24.34].x12 = 35/48 .x12

Thus, the middle term in the expansion of [3 – (x3/6)]7 are –105/8.x9 and 35/48.x12

1. Find the middle term in the expansion of (x/3 + 9y)10

Solution:

It is known that in the expansion (a+b)n, if n is even, then the middle term is (n/2 + 1)th term.

Therefore the middle term in the expansion of (x/3 + 9y)10 is (10/2 + 1)th term

Therefore, the middle term in the expansion of (x/3 + 9y)10 is (10/2 + 1)th = 6th

T6 = T5+1 = 10C5 (x/3)10-5 (9y)5 =10!/5!5! . 95.(x5/35).y5

= –10.9.8.7.6.5!/5.4.3.2.5! . [1/35].310.x5.y5

=252.35.x5.y5

Thus, the middle term in the expansion of (x/3 + 9y)10 is 61236x5y5

1. In the expansion of (1 + a)m+n prove that coefficient of am and an are equal

Solution:

It is known that (r+1)th term, (Tr+1) in the binomial expansion of (a+b)n is given by Tr+1 = nCr an-r br

Assuming that am occurs in the (r+1)th term, (Tr+1) in the binomial expansion of (a+b)n is given by Tr+1 = nCr an-r br

Assuming that am occurs in the (r+1)th term of the expansion (1+a)m+n, we obtain

Tr+1 = m+nCr(1)m+n-r(a)r = m+nCr.ar

Comparing the indices of a in am and in Tr+1, we obtain

r = m

Therefore the coefficient of am is

m+nCm = (m+n)!/m!(m+n-m)! = (m+n)!/m!n! ———–(1)

Assuming that an occurs in the (k+1)th term of the expansion (1 + a)m+n we obtain

Tk+1 = m+nCk(1)m+n-k(a)k = m+nCk(a)k

comparing th indices of a and an and in Tk+1

we obtain

k = n

herefore the coefficient of an is

m+nCn = (m+n)!/n!(m+n-m)! =(m+n)!/n!m!  ——————-(2)

Thus from (1) and(2) it can be observed that the coefficients of am and an in the expansion of (1+a)m+n are equal.

1. The coefficients of the (r-1)th, rth and (r+1)th terms in the expansion of (x+1)n are in the ratio 1:3:5. Find n and r.

Solution:

It is known that (k+1)th term, (Tk+1) in the binomial expansion of (a+b)n is given by Tk+1 = nCr an-k bk

Therefore (r – 1)th term in the expansion of (x+1)n is

Tr-1 = nCr-2 (x)n-(r-2)(1)r-2 = nCr-2 xn-r+2

(r+1) term in the expansion of (x+1)n is

Tr+1 = nCr(x)n-r(1)r = nCr xn-r

rth term in the expansion of (x + 1)n is

Tr = nCr -1 (x)n-(r-1)(1)(r-1) = nCr-1 xn-r+1

Therefore the coefficients of the (r -1)th, rth and (r +1)th terms in the expansion of (x + 1)n nCr -2, nCr-1 and nCr are respectively. Since these coefficients are in the ratio  1:3:5, we obtain

nCr-2/nCr-1  = 1/3 and nCr-1/nCr  = 3/5

nCr-2/nCr-1  = n!/(r-2)!(n-r+2)! x (r-1)(n-r+1)!/n! = (r-1)(r-2)(n-r+1)!/ !/(r-2)!(n-r+2)!(n-r+1)!

= r-1/n-r+2

r-1/n-r+2 = 1/3

3r – 3 = n – r + 2

n – 4r + 5 = 0

nCr-1/nCr  = n!/(r-1)!(n-r+1)! x (r!)(n-r)!/n! = r(r-1)!(n-r)!/(r-1)!(n-r+1)(n-r)!

=r/n-r+1

r/n-r+1 = 3/5

5r = 3n – 3r + 3

3n – 8r + 3 = 0

Multiplying (1) by 3 and subtracting it from (2), we obtain

4r – 12 = 0

⇒ r = 3

Putting the value of r in (1), we obtain n

– 12 + 5 = 0

⇒ n = 7

Thus, n = 7 and r = 3

1. Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1

Solution:

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

Tr+1 = 2nCr an-r br

Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain

Tr+1 = 2nCr (1)2n-r (x)r = 2nCr (x)r

Comparing the indices of x in xn and in Tr + 1, we obtain r = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n is

2nCn = (2n)!/n!(2n-n)! = (2n)!/n!n! = (2n)!/(n!)2———–(1)

Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2n – 1, we obtain

Tr+1 = 2n-1Ck (1)2n-1-k (x)k = 2n-1Ck (x)k

Comparing the indices of x in xn and Tk + 1, we obtain k = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n –1 is

2n-1Cn = (2n-1)!/n!(2n-1-n)! = (2n-1)!/n!(n-1)! )

= 2n.(2n-1)!/2n.n!(n-1)! = (2n)!/2.n!n! = 1/2[(2n)!/(n!)2] ————(2)

From (1) and (2) it is observed that

1/2 (2nCn) =2n-1Cn

2nCn =2(2n-1Cn)

Therefore,  the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.

Hence, proved.

1. Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Solution:

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

Tr+1 = 2nCr an-r br

Assuming that x2 occurs in the (r + 1)th term of the expansion of (1 + x)m, we obtain

Tr+1 = mCr (1)m-r (x)r = mCr (x)r

Comparing the indices of x in x2 and in Tr + 1, we obtain r = 2

Therefore, the coefficient of x2 is mC2 in the expansion of (1 + x)m is 6.

mC2 = 6

(m)!/2!(m-2)! =6

(m)(m-1)(m-2)!/2x(m-2)! = 6

m(m – 1) = 12

m2 – m – 12 = 0

m2 – 4m + 3m – 12 = 0

m(m-4)+3(m – 4) = 0

(m – 4)(m + 3) = 0

(m – 4) = 0 or (m + 3) = 0

m = 4 or m = -3

Thus, the positive value of m for which the coefficient of x2 in the expansion (1+ x)m is 6, is 4.