Binomial Theorems Class 11 OP Malhotra Exe-13B ISC Maths Solutions Ch-13 Latest editions. In this article you would learn about General, Middle and from the end Term. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Binomial Theorems Class 11 OP Malhotra Exe-13B ISC Maths Solutions Ch-13

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-13	Binomial Theorems
Writer	OP Malhotra
Exe-13(B)	General, Middle and from the end Term.

Exercise- 13B

Binomial Theorems Class 11 OP Malhotra Exe-13B Solution.

Que-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 = (1/3), b = (1/4).
(ii) Sixth term of [2x−(1/x²)]^7
(iii) Middle term of [2x−(1/y)]^8
(iv) Middle term of [(x^4)−(1/x³)]^11
(v) Middle term of [(x²/4)−(4/x²)]^10

Sol: (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 = 12!/(8!4!) × (2a)⁸ (3b)⁴ = {(12×11×10×9)/(4×3×2×1)} × 2⁸ a⁴ × 3⁴ b⁴ = 495 × 2⁸ × 3⁴ a⁴ b⁴
when a = 1/3  and  b = 1/4
∴ T₅ = 495 × 2⁸ × 3⁴ × (1/3)^4 × (1/4)^4
Thus T₅ = 495

(iv) On comparing [(x^4)−(1/x³)^11 with (x + a)^n; we have
‘x’ = x4; ‘a’ = –1/x³ and n = 11
Here no. of terms = n + 1 = 11 + 1 = 12
So there are two middle terms i.e.

Que-2: Find the term independent of x in the expansion of the following binomials :
(i) [x−(1/x)]^14
(ii) [(√x/3)−(√3/2x)]^12
(iii) [2x²−(1/x)]^12 what is its value?

Sol: (i) on comparing [x−(1/x)]^14 with (x + a)ⁿ; we have
‘x’ = x; ‘a’ = (-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 (-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 =14!/(7!7!)
= -(14×13×12×11×10×9×8)/(7×6×5×4×3×2) = -3432

Let (r + 1) th term be the term independent of x.
So putting exponent of x be equal to 0 .
i.e. (12−3r)/2 = 0 ⇒ r = 4
putting r = 4 in eqn. (1); we have
T₅ = 12C4 (1/3)^4 (-√3/2)^4
= 12!/(8!4!) {1/(3^4)} × {(3^2)/(2^4)}
= {(12×11×10×9)/(4×3×2×1)} × {1/3²×16)
= 5516

(iii) on comparing [2x²−(1/x)]^12 with (x + a)ⁿ; we have
‘x’ = 2x²; ‘a’ = (-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) (−1/x)^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 = 12!/(8!4!) × (2^4)
= {(12×11×10×9)/24} × 16
= 7920

Que-3: Find the coefficient of
(i) a⁶b³ in the expansion of [2a−(b/3)]^9
(ii) x⁷ in the expansion of [x²+(1/x)]^11
(iii) {1/(x^17)} in the expansion of [(x^4)−(1/x³)]^15
(iv) x⁴ in the expansion of [(x/2)−(3/x²)]^10

Sol: (i) comparing [2a−(b/3)]^9 with (x + a)^n ; we have
‘x’ = 2a; ‘a’ = –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) (-b/3)…(1)
Let T_{r + 1} contain term involving a⁶ b³
For this, we put r = 3
∴ from (1); T₄ = ⁹C₃ (2a)⁶ (-b/3)³ = {9!/(6!3!)} 2⁶ a⁶ × (-b³/3³)
= {(9×8×7)/6} × 64 × {(−1)³a^6 b³}/27 = −1792/9 a^6 b³
∴ Coeff. of a^6 b³ in the expansion of {2a−(b/3)}^9 be −1792/9.

(ii) Comparing (x^2+(1/x))^11 with (x + a)^n; we have ‘x’ = x^2; ‘a’ = 1/x and n = 11
We know that general term = Tr+1 = nCr xn-r dr
∴ Tr+1 = 11Cr (x^2)^11-r (1/x)^r = 11Cr x^(22-3r) …(1)
For term containing x^7, 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 (x^2+(1/x))^11 = 11C5 = 11!/5!6!
= (11×10×9×8×7)/(5×4×3×2 = 462

(iii) Comparing (x^4−(1/x^3))^15 with (x + a)^n; we have
‘x ‘ = x^4; ‘a’ = –1/x^3; n = 15
We know that, General term = Tr+1 = nCr x^(n-r) n^r
∴ Tr+1 = 15Cr (x^4)C^(15-r) (−1x/3)^r = 15Cr (-1)^r x^(60-7r) …(1)
Let Tr+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. 1/x^17 = 15C11 (-1)^11
= – 15!/(11!4!) = – (15×14×13×12)/24
= – 1365

(iv) Comparing [(x/2)−(3/x^2)^10 with (x + a)^n; we have
‘x’ = x/2; ‘a’ = −3/x^2 and n = 10
We know that, Tr+1 = nCr x^(n-r) d^r
= 10Cr(x/2)^(10−r) (−3/x^2)^r = 10Cr (1/2)^(10−r) (-3)^r x^(10-3r) ….(1)
For term containing x4 in the Expansion of [(x/2)−(3/x^2)]^10; we put 10 – 3r = 4 ⇒ r = 2
∴ from (1); T3 = 10C2 (1/2)^8 (-3)^2 x^4
∴ coeff of x^4 = 10C2 (1/2)^8 (-3)^2
= 10!/(8!2!) × (1/2^8) × 9
= {(10×9)/2} × (9/2^8) = 405256

Que-4: If the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are the same, find the value of a.

Sol: 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 ⇒ {9!/(2!7!)} 3. a² = {9!/(6!3!)} a³
⇒ {(9×8)/2} × 3 . a² = {(9×8×7)/6} a³
⇒ 108a² = 84a³ ⇒ 12 a² (7a – 9) = 0 ⇒ a = 0, 9/7
When a = 0 the given expansion no longer binomial.
∴ a = 9/7

Que-5: Write down the fourth term in the binomial expansion of [px+(1/x)]. 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 5/2

Sol: On comparing [px+(1/x)]^n with (x + a)ⁿ; we have
‘x’ = px; ‘a’ = (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 (1/x)^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 x^0
Also, given T4 = 5/2

p = 1/2

Que-6: The expansion by the binomial theorem of  [2x+(1/8)]^10  is 1024 x¹⁰ + 640x⁹ + ax⁸ + b x⁷ + ……. Calculate
(i) the numerical value of a and b;
(ii) coefficient of x⁸ in (3x – 2) [2x+(1/8)]^10
(iii) the value of x, for which the third and the fourth terms in the expansion of [2x+(1/8)]^10 are equal

Sol: On comparing [2x+(1/8)]^10 with (x + a)ⁿ; we have
‘x’ = 2x ; ‘a’ = 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 (1/8)^r
For term containing x⁸ we put r = 2 in eqn. (1); we get
T₃ = ¹⁰C₂(2x)⁸ (1/8)² = {10!/(8!2!)} (2^8) × (1/8²) × (x^8)
∴ Coeff. of x^8 = {10!/(8!2!)} × {(2^8)/(2^6)}
= {(10×9)/2} × 4 = 180Thus a = 180
For term containing x⁷, we put r = 3 in eqn. (1); we have
T₄ = ¹⁰C₃ (2x)^7 (1/8)³
∴ Coeff. of x⁷ = ¹⁰C₃ 2⁷ × = {10!/(7!3!)} × {(2^7)/(2^9)}
= ({(10×9×8)/6} × 14 = 30
Thus b = 30

(ii) Coefficient of x⁸; in (3x – 2) [2x+(1/8)]^10
= 3 × coeff. of x⁷ in [2x+(1/8)]^10  – 2 × Coeff. of x⁸ in [2x+(1/8)]^10
= 3 × 30 – 2 × 180 = 90 – 360 = – 270

(iii) For T₃ : putting r = 2 in eqn. (1); we have
T₃ = ¹⁰C₂ (2x)⁸ (1/8)²
For T₄ : putting r = 3 in eqn. (1); we have
T₄ = ¹⁰C₃ (2x)⁷ (1/8)³
according to given condition, T₃ = T₄

Que-7: Find the coefficient of x⁷ in [ax²+(1/bx)]^11 and the coefficient of [ax+(1/bx²)]^11. If these coefficients are equal, find the relation between a and b.

Sol: On comparing [ax²+(1/bx)]^11with (x + a)ⁿ; we have
‘x’ = ax²; ‘a’ = 1/bx n = 11
We know that general term = T_r+1 = ⁿC_r x^n-r a^r

For term containing x^-7 we put 11 – 3r = -7 ⇒ r = 6
Thus putting r = 6 in eqn. (3); we have

Que-8: In a binomial expansion, (x + a)ⁿ, the first three terms are 1, 56 and 1372 respectively. Find values of x and a.

Sol: 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); {n(n-1)/2} x^{n – 2}a² = 1372 …(7)
On dividing (3) by (5); we have (nx/2)= 56 …(8)
On dividing (7) by eqn. (6); we have
{(n−1)/2} (a/x) = 1372/56 = 49/2 …(9)
From (8) and (9); we have
{(n−1)/2} × (56/n) = 49/2
{(n−1)/n} = 49/56 = 7/8
⇒ 8n – 8 = 7n ⇒ n = 8
∴ from (5); x⁸ = 1 ⇒ x = 1
∴ from (8); {(8×a)/1}  = 56
⇒ a = 7

Que-9: Write the 4th term from the end in the expansion of [(x³/2)-(2/x³)]^9

Sol: On comparing [(x³/2)-(2/x³)]^9 with (x + a)ⁿ; we have
‘x’ = (x³/2) , ‘a’ = (-2/x²); 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
∴ Tr+1 = 9Cr (x³/2)^(9−r) (−2/x²)^r …(1)
For T7; Putting r = 6 in eqn. (1); we have
T7 = 9C6 (x³/2)³ (−2/x²)^6
= {9!/(6!3!)} {(x^9)/8} × {64/(x^12)}
= {(9×8×7)/6} × (8/x³)
= 672/x³

Que-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.

Sol: 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

Que-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) −3/10
(b) 10/3
(c) −5/3
(d) 3/5

Sol: 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
{4!/(2!2!)} α² = {–6!/(3!3!)} α³
⇒ 6α² = – {(6×5×4)/6} α³
⇒ 6a² + 20α³ = 0
⇒ 2α² (3 + 10α) = 0
⇒ α = 0,-3/10
Since α ≠ 0 if α = 0then given expansions be no longer binomial ∴ α = -3/10
∴ Ans. (a)

Que-12: 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}.

Sol: 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 T{(2n/2)+1} i.e. Tn+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

Que-13: Show that the middle term in the expansion of (1 + x)²ⁿ is [{1.3.5…(2n-1)}/n!].2^n x^n where n ∈ N.

Sol: 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 T[(2n/2)+1] i.e. Tn+1.
So putting r = n in eqn. (1); we have

Que-14: Find the coefficient of x⁵ in the expansion of 1 + (1 + x) + (1 + x)² + …. + (1 + x)¹⁰.

Sol: 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
= [{(1+x)^11−1}/{1+x−1}] = [{(1+x)^11−1}/x]
So coefficient of x⁵ in given expansion = Coeff. of x⁵ in [{(1+x)^11−1}/x]
= Coeff. of x⁶ in (1 + x)¹¹ – 1 = ¹¹C₆
= 11!/(6!5!)
= (11×10×9×8×7)/(5×4×3×2) = 462

Que-15: If x^p occurs in the expansion of [x²+(1/x)]^2n, prove that the coefficient is

Sol: On comparing [x²+(1/x)]^2n with (x + a)ⁿ we have, ‘x’ = x²; ‘a’ = 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 (1/x)^r ⇒ T_r+1 = ²ⁿC_r x^4n-2r (1/x^r) = ²ⁿC_r x^4n-3r …(1)
Let x^p occurs in (r + 1)th term of [x²+(1/x)]^2n
Therefore from eqn. (1); we have

Que-16: 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.

Sol: 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)

Que-17: 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.

Sol: 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.

Que-18: In the expansion of [x²+(1/x)]^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.

Sol: On comparing [x²+(1/x)]^n with (x + a)ⁿ; we have
‘ x’ = x²; ‘a’ = (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} (1/x)^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 = {11!/(5!6!)}x^7
= (11×10×9×8×7)/(5×4×3×2) x^7 = 462 x^7

Que-19: 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

Sol: By using binomial theorem, we have

Que-20: If rth and (r+1)th terms in the expression of (p+q)%n are equal, then prove that [{(n+1)q}/{r(p+q)}] = 1

Sol: Given, (p + q)ⁿ
∴ General term = T_{r + 1} = ⁿC_r p^{n – r} q^r
∴  rth term = T_r = ⁿC_r–1 (p)^{n – (r – 1)} q^{r – 1}
= ⁿC_{r – 1} p^{n – r + 1} q^{r – 1}
(r + 1)th term = T_r+1 = ⁿC_r p^{n – r} q^r
Given, T_r = T_{r + 1}
⇒ ⁿC_{r – 1} p^{n – r + 1} q^{r – 1} = ⁿC_r p^{n – r} q^r

Que-21: Find n in the binomial [√2+(1/√3)]^n, if the ratio of the 7th term from the beginning to the 7th term from the end is 1/6.

Sol: In the binomial expansion of  [√2+(1/√3)]^n [(n+1)-7+1]th  i.e., (n − 5)^th term from the beginning is the 7^th term from the end.
Now,

4 – (n/3) = 1
n = 9.

Que-22: 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.

Sol: 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

Que-23: (a) If the coefficients of second, third and fourth terms in the expression of (1+x)^2n are in A.P., then show that 2n²-9n+7 = 0

Sol: Given expression = (1 + x )²ⁿ
Coefficient of second term = ²ⁿC₁
Coefficient of third term = ²ⁿC₂
And coefficient of fourth term = ²ⁿC₃
As the given condition
²ⁿC₁. ²ⁿC₂ and ²ⁿC₃ are in A.P.
∴ ²ⁿC₂ – ²ⁿC₁ = ²ⁿC₃ – ²ⁿC₂
2. ²ⁿC₂ = ²ⁿC₁ + ²ⁿC₃

⇒ n(2n – 1) = n + [{n(2n-1)(2n-2)}/6]
⇒ 2n – 1 = 1 + [{(2n-1)(2n-2)}/6]
⇒ 12n – 6 = 6 + 4n² – 4n – 2n + 2
⇒ 12n – 12 = 4n² – 6n + 2
⇒ 4n² – 6n – 12n + 2 + 12 = 0
⇒ 4n² – 18n + 14 = 0
⇒ 2n² – 9n + 7 = 0
Hence proved.

(b) In the expression (1+x)^n, if the coefficients of 2nd, 3rd and 4th terms are in A.P., then find the value of n.

Sol: Given expression is (1 + x)ⁿ
(1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ⁿC₃x³ + … ⁿC_nxⁿ
Here, coefficient of 2^nd term = ⁿC₁
Coefficient of 3^rd term = ⁿC₂
And coefficient of 4^th term = ⁿC₃
Given that ⁿC₁, ⁿC₂ and ⁿC₃ are in A.P.
∴ 2 . ⁿC₂ = ⁿC₁ + ⁿC₃
⇒ 2 . [{n(n-1)}/2] = n + [{n(n-1)(n-2)}/3.2.1]
⇒ n(n-1) = n + [{n(n-1)(n-2)}/6]
⇒ n – 1 = 1 + [{(n-1)(n-2)}/6]
⇒ 6n – 6 = 6 + n² – 3n + 2
⇒ n² – 3n – 6n + 14 = 0
⇒ n² – 9n + 14 = 0
⇒ n² – 7n – 2n + 14 = 0
⇒ n(n – 7) – 2(n – 7) = 0
⇒ (n – 2)(n – 7) = 0
⇒ n = 2, 7
⇒ n = 7
Whereas n = 2 is not possible

–: End of Binomial Theorems Class 11 OP Malhotra Exe-13B ISC Maths Ch-13 Solutions. :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions

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