GSEB Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Find the coefficient of:
1. x⁵ in (x + 3)⁸
2. a⁵b⁷ in (a – 2b)¹²
Solutions to questions 1 and 2:
1. General term in (x + 3)⁸
= ⁸C_rx^8-r.3^r
We have to find the coefficient of x⁵
8 – r = 5 ⇒ r = 8 – 5 = 3.
Coefficient of x⁵ (putting r = 3)
= ⁸C₃.3³ = \(\frac{8.7.6}{1.2.3}\).27 = 56.27 = 1512.

2. (a – 2b)¹² = [a + (- 2b)]¹²
General term T_r+1 = C(12, r)a^12-r(- 2b)^r
Putting 12 – r = 5 or 12 – 5 = r ⇒ r = 7.
∴ T₇₊₁ = C(12, 7)a^12-7(- 2b)⁷
= C(12, 7)a⁵(- 2)⁷b⁷
= C(12, 7)(- 2)⁷a⁵ b⁷.
Hence, required coefficient is

= – 11 × 9 × 2⁷ = – 99 × 8 × 128
= – 101376.

3. General term = T_r+1 = ⁶C_r(x²)^6-r(- y)^r
= (- 1)^r\(\frac{6!}{r!(6 – r)!}\).x^12-2r.y^r.

4. Binomial expression is (x² – yx)¹².
∴ General term = T_r+1 = ¹²C_r(x²)^12-r.(- yx)^r
= \(\frac{12!}{r!(12 – r)!}\).x^24-2r(- 1)^ry^rx^r
= \(\frac{(-1)^{r} 12 !}{r !(12-r) !}\).x^24-ry^r.

5. Find the 4th term in the expansion of (x – 2y)¹².
Solution:
4th term = T₃₊₁ in the expansion of x + (- 2y)¹²
= ¹²C₃x^12-3(- 2y)³
= \(\frac{12.11.10}{1.2.3}\).x⁹(- 1)³.2³.y³
= – 220 × 8x⁹.y³ = – 1760x⁹y³.

6. Find the 13th term in the expansion of (9x – \(\frac{1}{3 \sqrt{x}}\)¹⁸)
Solution:
13th term = T₁₃ = T₁₂₊₁ = ¹⁸C₁₂(9x)^18-12(- \(\frac{1}{3 \sqrt{x}}\))¹²

Find the middle term in the expansion of:
7. (3 – \(\frac{x^{3}}{6}\))⁷
8. (\(\frac{x}{3}\) + 9y)¹⁰
Solutions to questions 7 and 8:
7. Number of terms in the expansion is 7 + 1 = 8.
∴ There are two middle terms which are T₄ and T₅.
Hence, we are to find T₄ and T₅ in the given expansion.
(3 – \(\frac{x^{3}}{6}\))⁷ = [3 + (- \(\frac{x^{3}}{6}\))⁷
∴ T_r+1 = C(7, 3)3^7-r(- \(\frac{x^{3}}{6}\))^r ………………… (1)
Now, T_r+1 = T₄.
or r + 1 = 4
∴ r = 3.
∴ Putting r = 3 in (1), we have:

Again T_r+1 = T₅ or r + 1 = 5.
⇒ r = 4
Putting r = 4 in (1), we get:

8. Number of terms in the expansion is 10 + 1 = 11.
Middle term of the expansion is \(\frac{11+1}{2}\) = T₆.
T_r+1 = C(10, r)(\(\frac{x}{3}\))^10-r (9r)^r.
But T_r+1 = T₆ or r + 1 = 6.
⇒ r = 5.
Putting r = 5 in (1), we have:

Question 9.
In the expansion of (1 + a)^m+n, prove that coefficients of a^m and aⁿ are equal?
Solution:
General term in the expansion of (1 + a)^m+n
= T_r+1 = ^m+nC_ra^r.
Putting r = m
T_m+1 = ^m+nC_ma^m ……………….. (1)
∴ Coefficient of a^m = ^m+nC_m.
Again putting r = n, we get
T_n+1 = ^m+nC_naⁿ.
Coefficients of aⁿ = ^m+nC_n
= ^m+nC_m ………………….. (2) [∵ ⁿC_r = ⁿC_n-r]
From (1) and (2), coefficient of am is equal to coefficient of
a^m is equal to coefficient of aⁿ.

Question 10.
The coefficients of the (r – 1)^th, rth and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio of 1 : 3 : 5. Find both n and r.
Solution:
General term in the expansion of (x + 1)ⁿ is
T_k+1 + 1 = C(n, k)x^n-k.
Putting T_k+1 = T_r-1 or k + 1 = r – 1 or k = r – 2.
∴ Coefficient of T_r-1 is C(n, r – 2) ………………….. (1)
Putting T_k+1 = T_r or k + 1 = r, we get
k = r – 1.
∴ Coefficient of T_r is C(n, r – 1) ………………….. (2)
Putting T_r+1 = T_k+1, we get
r = k.
∴ Coefficient of T_r+1 = C(n, r)
According to the problem,

or 5r = 3 (n + 1 – r)
or 8r = 3n + 3 ………………… (5)
From (4) and (5),
2n + 10 = 3n + 3
or 3n – 2n = 10 – 3
From (5), 8r = 21 + 3 = 24
∴ r = 3
∴ n = 7, r = 3.

Question 11.
Prove that the coefficient of xⁿ in (1 + x)²ⁿ  is twice the coefficient of xⁿ in (1 + x)^2n-1.
Solution:
General term in the expansion of (1 + x)²ⁿ is
T_r+1 = C(2n, r)x^r.
Putting r = n, we have:
T_n+1 = C(2n, n)xⁿ.
Coefficient of xⁿ = C(2n, n) ……………………. (1)
Again general term in the expansion of (1 + x)^2n-1 is
T_r+1 = C(2n – 1, r)x^r.
Putting r = n, we have:
T_n+1 = C(2n – 1, n)xⁿ.
Coefficient of xⁿ in the expansion of xⁿ is C(2n – 1, n).
According to the problem, we are to prove that
C(2n, n) = 2 × C(2n – 1, n).

Multiplying numerator (N) and denominator (D) by n on R.H.S; we have:
= \(\frac{2n(2n – 1)!}{n!n(n – 1)!}\).
i.e; \(\frac{2n!}{n!n!}\) = \(\frac{2n!}{n!n!}\), which is true. (Proved)

Question 12.
Find a positive value of m for which the coefficient of x² in the expansion of (1 + x)^m is 6.
Solution:
Coefficient of x² in the expansion of (1 + x)^m is C(m, 2).
According to the problem,
C(m, 2) = 6.
or \(\frac{m(m – 1)}{2!}\) = 6
or m² – m = 12
or m² – m – 12 = 0
or m² – 4m + 3m – 12 = 0
or m(m – 4) + 3(m – 4) = 0
or (m – 4)(m + 3) = 0
m = 4, since m ≠ – 3.