OP Malhotra Class 11 Maths Solutions Chapter 13 Binomial Theorem Ex 13(b)

Utilizing ISC Class 12 Maths Solutions OP Malhotra Chapter 13 Binomial Theorem Ex 13(b) as a study aid can enhance exam preparation.

S Chand Class 11 ICSE Maths Solutions Chapter 13 Binomial Theorem Ex 13(b)

Question 1.
Find the specified term of the expression in each of the following binomials :
(i) Fifth term of (2a + 3b)¹². Evaluate it when a = \(\frac { 1 }{ 3 }\), b = \(\frac { 1 }{ 4 }\).
(ii) Sixth term of \(\left(2 x-\frac{1}{x^2}\right)^7\)
(iii) Middle term of \(\left(2 x-\frac{1}{y}\right)^8\)
(iv) Middle term of \(\left(x^4-\frac{1}{x^3}\right)^{11}\)
(v) Middle term of \(\left(\frac{x^2}{4}-\frac{4}{x^2}\right)^{10}\)
Solution:
(i) On comparing (2a + 3b)¹² with (x + a)ⁿ
we have ‘x’ = 2a; ‘a’ = 3b; n = 12
We know that general term in the expansion of (x + a)ⁿ = T_r+1 = ⁿC_r x^{n – r} d^r
∴ general term in the expansion of (2a + 3b)¹² = ¹²C_r (2a)^12-r (3b)^r …(1)
For fifth term, putting r = 4 in eqn. (1); we have
∴ T₅ = ¹²C₄(2a)^{12 – 4} (3b)⁴ = \(\frac{12 !}{8 ! 4 !}\) × (2a)⁸ (3b)⁴ = \(\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}\) × 2⁸ a⁴ × 3⁴ b⁴ = 495 × 2⁸ × 3⁴ a⁴ b⁴
when a = \(\frac { 1 }{ 3 }\) and b = \(\frac { 1 }{ 4 }\)
∴ T₅ = 495 × 2⁸ × 3⁴ × \(\left(\frac{1}{3}\right)^4\) × \(\left(\frac{1}{4}\right)^4\)
Thus T₅ = 495

(iv) On comparing \(\left(x^4-\frac{1}{x^3}\right)^{11}\) with (x + a)ⁿ; we have
‘x’ = x⁴; ‘a’ = –\(\frac{1}{x^3}\) and n = 11
Here no. of terms = n + 1 = 11 + 1 = 12
So there are two middle terms i.e.

Question 2.
Find the term independent of x in the expansion of the following binomials :
(i) \(\left(x-\frac{1}{x}\right)^{14}\)
(ii) \(\left(\sqrt{\frac{x}{3}}-\frac{\sqrt{3}}{2 x}\right)^{12}\)
(iii) \(\left(2 x^2-\frac{1}{x}\right)^{12}\) what is its value?
Solution:
(i) on comparing \(\left(x-\frac{1}{x}\right)^{14}\) with (x + a)ⁿ; we have
‘x’ = x; ‘a’ = –\(\frac { 1 }{ x }\) and n = 14
We know that, general term = T_r+1 = ⁿC_n x^n-r d^r
i.e. T_r+1 = ¹⁴C_r x^14-r (-\(\frac { 1 }{ 4 }\))^r = ¹⁴C_r x^14-2r (-1)^r …(1)
Let T_r+1 be the term independent of x so we put exponent of x be equal to 0.
i.e. 14 – 2r = 0 ⇒ r = 7
Putting r = 7 in eqn. (1); we have
T₈ = ¹⁴C₇x⁰(-1)⁷ = \(\frac{14 !}{7 ! 7 !}\) = –\(\frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}{7 \times 6 \times 5 \times 4 \times 3 \times 2}\) = -3432

Let (r + 1) th term be the term independent of x.
So putting exponent of x be equal to 0 .
i.e. \(\frac{12-3 r}{2}\) = 0 ⇒ r = 4
putting r = 4 in eqn. (1); we have
T₅ = ¹²C₄\(\left(\frac{1}{3}\right)^4\) \(\left(\frac{-\sqrt{3}}{2}\right)^4\) = \(\frac{12 !}{8 ! 4 !}\) \(\frac{1}{3^4}\) × \(\frac{3^2}{2^4}\) = \(\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}\) × \(\frac{1}{3^2 \times 16}\) = \(\frac{55}{16}\)

(iii) on comparing \(\left(2 x^2-\frac{1}{x}\right)^{12}\) with (x + a)ⁿ; we have
‘x’ = 2x²; ‘a’ = – \(\frac { 1 }{ x }\) and n = 12
we know that, general term = T_r+1 = ⁿC_r x^n-r d^r
i.e. T_r+1 = ¹²C_r (2x²)^12-r\(\left(-\frac{1}{x}\right)^r\) = ¹²C_r 2^12-r (-1)^r x^24-2r-2 = ¹²C_r 2^12-r (-1)^r x^24-3r …(1)

Let T_r+1 be the term independent of x i.e. putting exponent of x be equal to 0.
i.e. 24 – 3r = 0 ⇒ r = 8
Putting r = 8 in eqn. (1); we have
T₉ = ¹²C₈ 2^12-8 (-1)⁸ = \(\frac{12 !}{8 ! 4 !}\) × 2⁴ = \(\frac{12 \times 11 \times 10 \times 9}{24}\) × 16 = 7920

Question 3.
Find the coefficient of
(i) a⁶b³ in the expansion of \(\left(2 a-\frac{b}{3}\right)^9\)
(ii) x⁷ in the expansion of \(\left(x^2+\frac{1}{x}\right)^{11}\)
(iii) \(\frac{1}{x^{17}}\) in the expansion of \(\left(x^4-\frac{1}{x^3}\right)^{15}\)
(iv) x⁴ in the expansion of \(\left(\frac{x}{2}-\frac{3}{x^2}\right)^{10}\)
Solution:
(i) comparing \(\left(2 a-\frac{b}{3}\right)^9\) with (x + a)ⁿ; we have
‘x’ = 2a; ‘a’ = –\(\frac { b }{ 3 }\); n = 9
We know that general term = T_{r + 1} = ⁿC_r x^n-r d^r
∴ T_{r + 1} = ⁹C_r (2a)^9-r \(\left(-\frac{b}{3}\right)^r\) …(1)
Let T_{r + 1} contain term involving a⁶ b³
For this, we put r = 3
∴ from (1); T₄ = ⁹C₃ (2a)⁶ \(\left(-\frac{b}{3}\right)^3\) = \(\frac{9 !}{6 ! 3 !}\) 2⁶ a⁶ × \(\frac{(-b)^3}{3^3}\)
= \(\frac{9 \times 8 \times 7}{6}\) × 64 × \(\frac{(-1)^3 a^6 b^3}{27}\) = \(\frac{-1792}{9}\) a⁶ b³
∴ Coeff. of a⁶b³ in the expansion of \(\left(2 a-\frac{b}{3}\right)^9\) be \(\frac{-1792}{9}\).

(ii) Comparing \(\left(x^2+\frac{1}{x}\right)^{11}\) with (x + a)ⁿ; we have ‘x’ = x²; ‘a’ = \(\frac{1}{x}\) and n = 11
We know that general term = T_r+1 = ⁿC_r x^n-r d^r
∴ T_r+1 = ¹¹C_r (x²)^11-r \(\left(\frac{1}{x}\right)^r\) = ¹¹C_r x^22-3r …(1)
For term containing x⁷, we put 22 – 3r = 7 ⇒ r = 5
putting r = 5 in eqn. (1); we have
T₆ = ¹¹C₅ x⁷
∴ Coeff. of x⁷ in the expansion of \(\left(x^2+\frac{1}{x}\right)^{11}\) = ¹¹C₅ = \(\frac{11 !}{5 ! 6 !}\) = \(\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2}\) = 462

(iii) Comparing \(\left(x^4-\frac{1}{x^3}\right)^{15}\) with (x + a)ⁿ; we have
‘x ‘ = x⁴; ‘a’ = –\(\frac{1}{x^3}\); n = 15
We know that, General term = T_r+1 = ⁿC_r x^n-r n^r
∴ T_r+1 = ¹⁵C_r (x⁴)C^15-r \(\left(\frac{-1}{x^3}\right)^r\) = ¹⁵C_r (-1)^r x^60-7r …(1)
Let T_r+1 containing term involving x^-17
For this we put 60 – 7r = – 17 ⇒ r = 11
putting r = 11 in eqn. (1); we have
∴ T₁₂ = ¹⁵C₁₁ (-1)¹¹ x^-17
∴ coeff. of x^-17 i.e. \(\frac{1}{x^{17}}\) = ¹⁵C₁₁ (-1)¹¹ = – \(\frac{15 !}{11 ! 4 !}\) = – \(\frac{15 \times 14 \times 13 \times 12}{24}\) = – 1365

(iv) Comparing \(\left(\frac{x}{2}-\frac{3}{x^2}\right)^{10}\) with (x + a)ⁿ; we have
‘x’ = \(\frac{x}{2}\); ‘a’ = \(\frac{-3}{x^2}\) and n = 10
We know that, T_r+1 = ⁿC_r x^n-r d^r
= ¹⁰C_r\(\left(\frac{x}{2}\right)^{10-r}\) \(\left(\frac{-3}{x^2}\right)^r\) = ¹⁰C_r \(\left(\frac{1}{2}\right)^{10-r}\) (-3)^r x^10-3r ….(1)
For term containing x⁴ in the Expansion of \(\left(\frac{x}{2}-\frac{3}{x^2}\right)^{10}\); we put 10 – 3r = 4 ⇒ r = 2
∴ from (1); T₃ = ¹⁰C₂ \(\left(\frac{1}{2}\right)^8\) (-3)² x⁴
∴ coeff of x⁴ = ¹⁰C₂ \(\left(\frac{1}{2}\right)^8\) (-3)² = \(\frac{10 !}{8 ! 2 !}\) × \(\frac{1}{2^8}\) × 9 = \(\frac{10 \times 9}{2}\) × \(\frac{9}{2^8}\) = \(\frac{405}{256}\)

Question 4.
If the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are the same, find the value of a.
Solution:
On comparing (3 + ax)⁹ with (x + a)ⁿ; we have
‘ x’ = 3; ‘a’ = ax and n = 9
We know that, General Term = T_{r + 1} = ⁿC_r x^n-r d^r
∴ T_{r + 1} = ⁹C_r 3^9-r (ax)^r = ⁹C_r 3^9-r d^r x^r
For term containing x², we put r = 2 in eqn. (1); T₃ = ⁹C₂ 3⁷ a² x²
∴ Coeff. of x² = ⁹C₂ 3⁷ . a²
For term containing x³, we put r = 3 in eqn. (1) ; T₄ = ⁹C₃ 3⁶ a³ x³
∴ Coeff. of x³ = ⁹C₃ 3⁶ . a³
According to given condition, we have
⁹C₂ 3⁷ . a² = ⁹C₃ 3⁶ . a³ ⇒ \(\frac{9 !}{7 ! 2 !}\) 3. a² = \(\frac{9 !}{6 ! 3 !}\) a³
⇒ \(\frac{9 \times 8}{2}\) × 3 . a² = \(\frac{9 \times 8 \times 7}{6}\) a³ ⇒ 108a² = 84a³ ⇒ 12 a² (7a – 9) = 0 ⇒ a = 0, \(\frac{9}{7}\)
When a = 0 the given expansion no longer binomial.
∴ a = \(\frac{9}{7}\)

Question 5.
Write down the fourth term in the binomial expansion of \(\left(p x+\frac{1}{x}\right)^n\). If this term is independent of x, find the value of n. With this value of n, calculate the value of p given that the fourth term is equal to \(\frac{5}{2}\).
Solution:
On comparing \(\left(p x+\frac{1}{x}\right)^n\) with (x + a)ⁿ; we have
‘x’ = px; ‘a’ = \(\frac{1}{x}\); n = n
We know that, general term = T_r+1 = ⁿC_r x^n-r d^r
⇒ T_r+1 = ⁿC_r (px)^n-r \(\left(\frac{1}{x}\right)^r\) = ⁿC_r P^n-r x^n-2r …(1)
For T₄; we put r = 3 in eqn. (1); we get
T₄ = ⁿC₃ P^{n – 3} x^{n – 6} …(2)
For term independent of x, we put n – 6 = 0 ⇒ n = 6
∴ from (2); T₄ = ⁶C₃ P³ x⁰

Question 6.
The expansion by the binomial theorem of \(\left(2 x+\frac{1}{8}\right)^{10}\) is 1024 x¹⁰ + 640x⁹ + ax⁸ + b x⁷ + ……. Calculate
(i) the numerical value of a and b;
(ii) coefficient of x⁸ in (3x – 2)\(\left(2 x+\frac{1}{8}\right)^{10}\);
(iii) the value of x, for which the third and the fourth terms in the expansion of \(\left(2 x+\frac{1}{8}\right)^{10}\) are equal.
Solution:
On comparing \(\left(2 x+\frac{1}{8}\right)^{10}\) with (x + a)ⁿ; we have
‘x’ = 2x ; ‘a’ = \(\frac { 1 }{ 8 }\) and n = 10
We know that, general term = T_r+1 = ⁿC_r x^n-r d^r
∴ T^r+1 = ¹⁰C_r (2x)^10-r \(\left(\frac{1}{8}\right)^r\)
For term containing x⁸ we put r = 2 in eqn. (1); we get

T₃ = ¹⁰C₂(2x)⁸ \(\left(\frac{1}{8}\right)^2\) = \(\frac{10 !}{8 ! 2 !}\) 2⁸ × \(\frac{1}{8^2}\) × x⁸
∴ Coeff. of x⁸ = \(\frac{10 !}{8 ! 2 !}\) × \(\frac{2^8}{2^6}\) = \(\frac{10 \times 9}{2}\) × 4 = 180
Thus a = 180
For term containing x⁷, we put r = 3 in eqn. (1); we have
T₄ = ¹⁰C₃ (2x)⁷ \(\left(\frac{1}{8}\right)^3\)
∴ Coeff. of x⁷ = ¹⁰C₃ 2⁷ × \(\frac{1}{8^3}\) = \(\frac{10 !}{7 ! 3 !}\) × \(\frac{2^7}{2^9}\) = \(\frac{10 \times 9 \times 8}{6}\) × \(\frac{1}{4}\) = 30
Thus b = 30

(ii) Coefficient of x⁸; in (3x – 2) \(\left(2 x+\frac{1}{8}\right)^{10}\)
= 3 × coeff. of x⁷ in \(\left(2 x+\frac{1}{8}\right)^{10}\) – 2 × Coeff. of x⁸ in \(\left(2 x+\frac{1}{8}\right)^{10}\)
= 3 × 30 – 2 × 180 = 90 – 360 = – 270

(iii) For T₃ : putting r = 2 in eqn. (1); we have
T₃ = ¹⁰C₂ (2x)⁸ \(\left(\frac{1}{8}\right)^2\)
For T₄ : putting r = 3 in eqn. (1); we have
T₄ = ¹⁰C₃ (2x)⁷ \(\left(\frac{1}{8}\right)^3\)
according to given condition, T₃ = T₄

Question 7.
Find the coefficient of x⁷ in \(\left(a x^2+\frac{1}{b x}\right)^{11}\) and the coefficient of \(\left(a x+\frac{1}{b x^2}\right)^{11}\). If these coefficients are equal, find the relation between a and b.
Solution:
On comparing \(\left(a x^2+\frac{1}{b x}\right)^{11}\) with (x + a)ⁿ; we have
‘x’ = ax²; ‘a’ = \(\frac { 1 }{ bx }\); n = 11
We know that general term = T_r+1 = ⁿC_r x^n-r a^r
i.e. T_r+1 = ¹¹C_r (ax²)^11-r \(\left(\frac{1}{b x}\right)^r\) = ¹¹C_r a^11-r \(\frac{1}{b^r}\) x_22-3r …(1)
For term containing x⁷ we put 22 – 3r = 7 ⇒ r = 5
putting r = 5 in eqn. (1); we have
T₆ = ¹¹C₅ a⁶\(\frac{1}{b^5}\) x⁷
This coefficient of x⁷ in \(\left(a x^2+\frac{1}{b x}\right)^{11}\) = ¹¹C₅\(\frac{a^6}{b^5}\) …(2)
on comparing \(\left(a x+\frac{1}{b x^2}\right)^{11}\) with (x + a)ⁿ
we have ‘x’ = ax; ‘a’ = \(\frac{1}{b x^2}\) and n = 11
∴ General term = T_r+1 = ⁿC_r x^n-r a^r = ¹¹C_r (ax)^{11 – r}\(\left(\frac{1}{b x^2}\right)^r\) = ¹¹C_r\(\frac{a^{11-r}}{b^r}\) x^11-3r …(3)
For term containing x^-7 we put 11 – 3r = -7 ⇒ r = 6
Thus putting r = 6 in eqn. (3); we have

Question 8.
In a binomial expansion, (x + a)ⁿ, the first three terms are 1, 56 and 1372 respectively. Find values of x and a.
Solution:
General term in the expansion of (x + a)ⁿ = T_r+1 = ⁿC_r x^n-r a^r …(1)
put r = 0 in eqn. (1); T₁ = ⁿC₀ xⁿ a⁰ = 1 …(2)
put r = 1 in eqn. (2); T₂ = ⁿC₁ x^n-1 a = 56 …(3)
put r = 2 in eqn. (1); T₃ = ⁿC₂ x^n-2 a² = 1372 …(4)
∴ from (2); xⁿ = 1 …(5)
from (3); nx^{n – 1} a = 56 …(6)
from (4); \(\frac{n(n-1)}{2}\) x^{n – 2}a² = 1372 …(7)
On dividing (3) by (5); we have \(\frac{na}{x}\) = 56 …(8)
On dividing (7) by eqn. (6); we have
\(\frac{(n-1)}{2}\)\(\frac{a}{x}\) = \(\frac{1372}{56}\) = \(\frac{49}{2}\) …(9)
From (8) and (9); we have
\(\frac{n-1}{2}\) × \(\frac{56}{n}\) = \(\frac{49}{2}\)
\(\frac{n-1}{n}\) = \(\frac{49}{56}\) = \(\frac{7}{8}\)
⇒ 8n – 8 = 7n ⇒ n = 8
∴ from (5); x⁸ = 1 ⇒ x = 1
∴ from (8); \(\frac{8 \times a}{1}\) = 56 ⇒ a = 7

Question 9.
Write the 4th term from the end in the expansion of \(\left(\frac{x^3}{2}-\frac{2}{x^2}\right)^9\).
Solution:
On comparing \(\left(\frac{x^3}{2}-\frac{2}{x^2}\right)^9\) with (x + a)ⁿ; we have
‘x’ = \(\frac{x^3}{2}\), ‘a’ = \(\frac{-2}{x^2}\); n = 9
We know that, rth term from end = (n – r + 2)th term from beginning
i.e. 4th term from end = (9 – 4 + 2)th i.e. 7 th term from beginning
We know that, general term = T_r+1 = ⁿC_r x^n-r a¹
∴ T_r+1 = ⁹C_r\(\left(\frac{x^3}{2}\right)^{9-r}\) \(\left(\frac{-2}{x^2}\right)^r\) …(1)
For T₇; Putting r = 6 in eqn. (1); we have
T₇ = ⁹C₆\(\left(\frac{x^3}{2}\right)^3\) \(\left(\frac{-2}{x^2}\right)^6\) = \(\frac{9 !}{6 ! 3 !}\) \(\frac{x^9}{8}\) × \(\frac{64}{x^{12}}\) = \(\frac{9 \times 8 \times 7}{6}\) × \(\frac{8}{x^3}\) = \(\frac{672}{x^3}\)

Question 10.
The coefficients of (2r + 1)th and (r + 2)th terms in the expansions of (1 + x)⁴³ are equal. Find the value of r.
Solution:
On comparing (1 + x)⁴³ with (x + a)ⁿ; we have
‘x’ = 1; ‘a’ = x; n = 43
We know that general term = T_r+1 = ⁿC_r x^n-r a^r
⇒ T_r+1 = ⁴³C_r 1^43-r x^r = ⁴³C_r x^r …(1)
For T_{(2 r+1)}; we put r = ‘2r ‘ in eqn. (1) ; T_2r+1 = ⁴³C_2r x^2r
∴ Coeff. of T_{2 r+1} = ⁴³C_2r
For T_r+2; we replace r by r + 1 in eqn. (1); we have
T_r+2 = ⁴³C_r+1 x^r+1
∴ Coeff. of T_r+2 = ⁴³C_r+1
According to given condition, ⁴³C_2r = ⁴³C_r+1
2r = r + 1 or 2r + r + 1 = 43 ⇒ r = 1 or r = 14

Question 11.
The coefficient of the middle term in the binomial expansion in powers of x of (1 + αx)⁴ and of (1 – αx)⁶ is the same if α equals
(a) \(\frac{-3}{10}\)
(b) \(\frac{10}{3}\)
(c) \(\frac{-5}{3}\)
(d) \(\frac{3}{5}\)
Solution:
On comparing (1 + αx)⁴ with (x + a)ⁿ; we have
‘x’ = 1 ; ‘a’ = αx; n = 4
∴ General term = T_r+1 = ⁿC_r x^n-r a^r = ⁴C_r 1^4-r (αx)^r
∴ T_r+1 = ⁴C_r α^r x^r
Here no. of terms = n + 1 = 4 + 1 = 5 (odd)
∴ there is only one middle term and is given by

For T₃: putting r = 2 in eqn. (1); we have
T₃ = ⁴C₂ α² x²
∴ Coeff. of middle term T₃ = ⁴C₂ α²
On comparing (1 – αx)⁶ with (x + a)ⁿ; we have
‘x’ = 1; ‘a’ = – αx; n = 6
∴ no. of terms = n + 1 = 7 (odd). So there is only one middle term and is given by

For T₄: putting r = 3 in eqn. (2); we have
T₃ = –⁶C₃ α³ x³
∴ Coeff. of middle term T₄ = –⁶C₃ α³
According to given condition; ⁴C₂ α² = –⁶C₃ α³
\(\frac{4 !}{2 ! 2 !} \alpha^2\) = –\(\frac{6 !}{3 ! 3 !} \alpha^3\) ⇒ 6α² = – \(\frac{6 \times 5 \times 4}{6}\) α³
⇒ 6a² + 20α³ = 0
⇒ 2α² (3 + 10α) = 0
⇒ α = 0,\(\frac{-3}{10}\)
Since α ≠ 0 if α = 0then given expansions be no longer binomial ∴ α = \(\frac{-3}{10}\)
∴ Ans. (a)

Question 12.
Find the sixth term of the expansion of (y^1/2 + x^1/3)ⁿ, if the binomial coefficient of the third term from the end is 45.
Solution;
On comparing (y^1/2 + x^1/3)ⁿ with (x + a)ⁿ; we have
‘x’ = y^1/2; ‘a’ = x^1/3; n = n
We know that, general term = T_r+1 = ⁿC_r x^n-r α^r
∴ T_r+1 = ⁿC_r (y^1/2)^n-r (x^1/3)^r …(1)
For T₆; putting r = 5 in eqn. (1); we have
T₆ = ⁿC₅ (y^1/2)^n-5 (x^1/3)⁵ …(2)
We know that, rth term from end in (x + a)ⁿ = (n – r + 2)th term from beginning in (x + a)ⁿ
Thus 3rd term from end = (n – 3 + 2) th i.e. (n – 1) th term from beginning
Replacing r by n – 2 in eqn. (1); we have
T_{n – 1} = ⁿC_{n – 2} (y^1/2)^{n – (n -2)} (x^1/3)^n-2 = ⁿC₂ (y^1/2)² (x^1/3)^{n – 2} [∵ ⁿC_r = ⁿC_{n – r}]
given coeff. of 3rd term from end = 45

Question 13.
Show that the coefficient of the middle term in the expansion of (1 + x)²ⁿ is the sum of the coefficients of two middle terms in the expansion of (1 + x)^{2n – 1}.
Solution:
On comparing (1 + x)²ⁿ with (x + a)ⁿ; we have
‘x’ = 1 ; ‘a’ = x; ‘n’ = 2n
We know that, general term = T_r+1 = ⁿC^r x^{n – r} a^r
∴ T_r+1 = ²ⁿC_r 1^{2n – r} x^r = ²ⁿC_r x^r …(1)
Here no. of terms = 2n + 1 (odd) ; so their is only one middle term given by \(\mathrm{T}_{\frac{2 n}{2}+1} \text { i.e. } \mathrm{T}_{n+1}\)
putting r = n in eqn. (1); we have
T_n+1 = ²ⁿC_n xⁿ
Therefore, coeff. of middle term in (1 + x)²ⁿ = ²ⁿC_n
On comparing (1 + x)^{2n – 1} with (x + a)ⁿ; ‘x’ = 1; ‘a’ = x ; ‘n’ = 2n – 1
∴ no. of terms = 2n – 1 + 1 = 2n (even)
Thus there are two middle terms given by

Question 14.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1.3 .5 \ldots(2 n-1)}{n !} .2^n x^n\) where n ∈ N.
Solution:
On comparing (1 + x)²ⁿ with (x + a)ⁿ; we have
‘x’ = 1 ;’a’ = x ; ‘n’ = 2n
We know that, general term = T_r+1 = ⁿC_r x^n-r a^r
∴ T_r+1 = ²ⁿC_r 1^2n-r x^r = ²ⁿC_r x^r …(1)
Here no. of terms = 2n + 1 = odd
So there is only one middle term given by \(\mathrm{T}_{\frac{2 n}{2}+1} \text { i.e. } \mathrm{T}_{n+1} \text {. }\)
So putting r = n in eqn. (1); we have

Question 15.
Find the coefficient of x⁵ in the expansion of 1 + (1 + x) + (1 + x)² + …. + (1 + x)¹⁰.
Solution:
Given expansion = 1 + (1 + x) + (1 + x)² …. + (1 + x)¹⁰
it form G.P with first term = 1 and common ratio = 1 + x > 1 and n = 11
= \(\frac{(1+x)^{11}-1}{1+x-1}\) = \(\frac{(1+x)^{11}-1}{x}\)
So coefficient of x⁵ in given expansion = Coeff. of x⁵ in \(\left\{\frac{(1+x)^{11}-1}{x}\right\}\)
= Coeff. of x⁶ in (1 + x)¹¹ – 1 = ¹¹C₆
[∵ General term = T_r+1 = ¹¹C_r 1^{11 – r} x^r]
= \(\frac{11 !}{6 ! 5 !}\) = \(\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2}\) = 462

Question 16.
If x^p occurs in the expansion of \(\left(x^2+\frac{1}{x}\right)^{2 n}\), prove that the coefficient is

Solution:
On comparing \(\left(x^2+\frac{1}{x}\right)^{2 n}\)with (x + a)ⁿ we have, ‘x’ = x²; ‘a’ = \(\frac{1}{x}\); ‘n’ = 2n
We know that, General term = T_r+1 = ⁿC_r x^n-r a^r
∴ T_r+1 = ²ⁿC_r(x²)^2n-r \(\left(\frac{1}{x}\right)^r\) ⇒ T_r+1 = ²ⁿC_r x^4n-2r \(\frac{1}{x^r}\) = ²ⁿC_r x^4n-3r …(1)
Let x^p occurs in (r + 1)th term of \(\left(x^2+\frac{1}{x}\right)^{2 n}\)
Therefore from eqn. (1); we have

Question 17.
If P be the sum of odd terms and Q be the sum of even terms in the expansion of (x + a)ⁿ, prove that
(i) P² – Q² = (x² – a²)ⁿ,
(ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ and
(iii) 2 (P² + Q²) = (x + a)²ⁿ + (x – a)²ⁿ.
Solution:
We know that, general term in the expansion of (x + a)ⁿ = T_r+1 = ⁿC_r x^n-r a^r
∴ (x + a)ⁿ = T₁ + T₂ + T₃ + ….
⇒ (x + a)ⁿ = (T₁ + T₃ + T₅ + ….) + (T₂ + T₄ + T₆ + ….) = P + Q
⇒ (x + a)ⁿ = P + Q …(1)
Also, (x – a)ⁿ = T₁ – T₂ + T₃ – T₄ + …… = (T₁ + T₃ + T₅ + …..) – (T₂ + T₄ + T₆ + ….)
⇒ (x – a)ⁿ = P – Q …(2)
(i) P² – Q² = (P – Q)(P + Q) = (x – a)ⁿ (x + a)ⁿ = [x² – a²]ⁿ

(ii) (x + a)²ⁿ = (P + Q)² …(3)
and (x – a)²ⁿ = (P – Q)² …(4)
Thus, (x + a)²ⁿ – (x – a)²ⁿ = (P + Q)² – (P – Q)² = P² + Q² + 2PQ – P² – Q² + 2PQ = 4PQ

(iii) On adding (3) and (4); we have
(x+a)²ⁿ + (x – a)²ⁿ = (P + Q)² + (P – Q)² = 2(P² + Q²)

Question 18.
If the coefficients of the rth, (r + 1)th and (r + 2)th terms in the expansion of (1 + x)ⁿ are in A.P., prove that n² – n(4r + 1) + 4r² – 2 = 0.
Solution:
We know that, general term = T_r+1 = ⁿC_r x^n-r a^r
[Here x = 1 ; a = x ; n = n]
∴ T_r+1 = ⁿC_r 1^n-r x^r = ⁿC_r x^r
∴ Coeff. of r th term = ⁿC_{r – 1} and Coeff. of (r + 2)th term = ⁿC_r+1
Coeff. of (r + 1)th term = ⁿC_r
Since coefficients of rth, (r + 1)th and (r + 2)th terms in the expansion of (1 + x)ⁿ are in A.P.

Question 19.
In the expansion of \(\left(x^2+\frac{1}{x}\right)^n\), the coefficient of the fourth term is equal to the coefficient of the ninth term. Find n and the sixth term of the expansion.
Solution:
On comparing \(\left(x^2+\frac{1}{x}\right)^n\) with (x + a)ⁿ; we have
‘ x’ = x²; ‘a’ = \(\frac { 1 }{ x }\); n = n
We know that, General term = T_r+1 = ⁿC_r x^n-r a^r
⇒ T_r+1 = ⁿC_r (x²)^{n – r} \(\left(\frac{1}{x}\right)^r\) = ⁿC_r x^{2n – 3r} …(1)
For T₄; putting r = 3 in eqn. (1); we have
T₄ = ⁿC₃ x^2n-9
For T₉; putting r = 8 in eqn. (1); we have
T₉ = ⁿC₈ x^2n-24
Given coeff. of T₄ = coeff of T₉
⇒ ⁿC₃ = ⁿC₈ ⇒ n = 3 + 8 = 11
[∵ ⁿC_r = ⁿC_s ⇒ r = s or r + s = n]
For T₆; putting r = 5 in eqn. (1); we have
∴ T₆ = ¹¹C₅ x^{22 – 15} =¹¹C₅ x⁷ = \(\frac{11 !}{5 ! 6 !}\)x⁷ = \(\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2}\) x⁷ = 462 x⁷

Question 20.
The coefficient of xⁿ in the expansion of (1 + x)(1 – x)ⁿ is
(a) (- 1)^n-1 . (n – 1)²
(b) (-1)ⁿ (1 – n)
(c) n – 1
(d) (- 1)^n-1.n
Solution:
By using binomial theorem, we have