Students can find the JEE Advanced Maths binomial theorem, previous years’ questions and solutions on this page. Expansion of binomial, general term, coefficient of any power of x, properties of binomial coefficients, number of terms in an expansion, etc., are the important topics in binomial theorem. Practising the JEE Advanced Maths binomial theorem questions and solutions will help students become familiar with the type of questions asked from this topic.

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Question 1: Coefficient of x¹¹ in the expansion (1 + x²)⁴ (1 + x³)⁷ (1 + x⁴)¹² is

(a) 1051

(b) 1106

(c) 1113

(d) 1120

Solution:

(1 + x²)⁴ (1 + x³)⁷ (1 + x⁴)¹²

=> coefficient of x^a × coefficient of x^b × coefficient of x^c

So that a+b+c = 1

Here a = 2m, b = 3n, c = 4p

So 2m + 3n + 4p = 11

Possibilities are m = 0, n = 1, p = 2

m = 1, n = 3, p = 0

m = 2, n = 1, p = 1

m = 4, n = 1, p = 0

Required coefficient = ⁴C₀×⁷C₁×¹²C₂ + ⁴C₁×⁷C₃×¹²C₀ + ⁴C₂×⁷C₁×¹²C₁ + ⁴C₄×⁷C₁×¹²C₀

= 462 + 140 + 504 + 7

= 1113

Hence option c is the answer.

Question 2: The coefficients of three consecutive terms of (1+x)ⁿ⁺⁵ are in the ratio 5: 10: 14. Then n =

Solution:

T_r+1 = ⁿ⁺⁵C_rx^r

Given ⁿ⁺⁵C_r-1 : ⁿ⁺⁵C_r : ⁿ⁺⁵C_r+1 = 5: 10: 14

ⁿ⁺⁵C_r/ ⁿ⁺⁵C_r-1 = 10/5

=> (n+5-r+1)/r = 2

=> (n+6-r)/r = 2

=> n+6-r = 2r

n = 3r-6 ..(i)

ⁿ⁺⁵C_r+1/ ⁿ⁺⁵C_r = 14/10

=> ((n+5)-(r+1)+1)/(r+1) = 7/5

=> (n+5-r)/(r+1) = 7/5

=> 5n+25-5r = 7r + 7

=> 5n – 12r + 18 = 0 ..(ii)

Put (i) in (ii)

5(3r-6) – 12r + 18 = 0

=> 15r – 30 – 12r + 18 = 0

=> 3r – 12 = 0

=> r = 4

So n = 3r-6

= 12-6

= 6

Hence the value of n is 6.

Question 3: The coefficient of x⁹ in the expansion (1+x)(1+x²)(1+x³) ..(1+x¹⁰⁰) is

Solution:

Given expansion (1+x)(1+x²)(1+x³) ..(1+x¹⁰⁰)

To get power 9, the possibilities are

(0, 9), (1, 8), (2, 7), (3, 6), (4, 5) = 5 ways

Also (1, 2, 6), (1, 3, 5), (2, 3, 5) = 3 ways

So the coefficient of x⁹ in the given expression is 8.

Question 4: If {p} denotes the fractional part of the number p, then {3²⁰⁰/8}, is equal to

(a) 5/8

(b) ⅞

(c) ⅜

(d) 1/8

Solution:

3²⁰⁰/8 = (⅛)9¹⁰⁰

= ⅛ (1 + 8)¹⁰⁰

= ⅛ (1 + ¹⁰⁰C₁.8 + ¹⁰⁰C₂.8² + …+¹⁰⁰C₁₀₀.8¹⁰⁰)

= 1/8 + integer

So {1/8 + integer} = 1/8

Hence option d is the answer.

Question 5: The coefficient of x⁷ in the expression (1 + x)¹⁰ + x(1 + x)⁹ + x²(1 + x)⁸ + ⋯ + x¹⁰ is :

(a) 420

(b) 330

(c) 210

(d) 120

Solution:

Let S= (1+x)¹⁰+ x(1+x)⁹+ x²(1+x)⁸+⋯ +x¹⁰

Applying sum of terms of G.P

S = (1+x)¹⁰(1 – (x/(1+x))¹¹ )/(1 – x/(1+x))

= (1 + x)¹¹ – x¹¹

Coefficient of x⁷ is ¹¹C₇ = 11!/7! 4!

= 330

Hence option b is the answer.

Question 6: The coefficient of x⁴ in the expansion of (1+x+x²)¹⁰ is

Solution:

General term of the given expression is given by (10!/p!q!r!)x^q+2r

Here, q+2r = 4

When p = 6, q = 4, r = 0, coefficient = 10!/(6!×4!) = 210

When p = 7, q = 2, r = 1, coefficient = 10!/(7!×2!×1!) = 360

When p = 8, q = 0, r = 2, coefficient = 10!/(8!×2!) = 45

Sum = 615

Question 7: If C_r = ²⁵C_r and C₀+5⋅C₁+9⋅C₂ +⋯+101. C₂₅ =2²⁵.k then k is equal to

Solution:

S = ²⁵C₀ + 5 ²⁵C₁ + 9 ²⁵C₂ + ….+97 ²⁵C₂₄ +101 ²⁵C₂₅ = 2²⁵k ..(i)

Reverse and apply property ⁿC_r = ⁿC_n-r in all coefficients

S = 101 ²⁵C₀ + 97 ²⁵C₁+…+5 ²⁵C₂₄+²⁵C₂₅ …(ii)

Adding (i) and (ii)

2S = 102 [²⁵C₀+²⁵C₁+…+²⁵C₅]

S = 51×2²⁵

k = 51

Question 8: If the fractional part of the number 2⁴⁰³/15 is k/15, then k is equal to:

(a) 4

(b) 8

(c) 6

(d) 14

Solution:

2⁴⁰³ = 2⁴⁰⁰ × 2³

= (2⁴)¹⁰⁰×8

= 8×16¹⁰⁰

= 8(1 + 15)¹⁰⁰

When 2⁴⁰³ is divided by 15

=> (8/15)(1+15x) = 8x + 8/15

So fractional part of the number is 8/15.

So k = 8

Hence option b is the answer.

Question 9: The number of terms in the expansion of the (1+x)¹⁰¹(1+x² -x)¹⁰⁰ in powers of x is

(a) 302

(b) 301

(c) 202

(d) 101

Solution:

(1+x)¹⁰¹(1+x² -x)¹⁰⁰ = (1+x)(1+x)¹⁰⁰(1-x+x²)¹⁰⁰

= (1+x)[(1+x)(1-x+x²)]¹⁰⁰

= (1+x)(1+x³)¹⁰⁰

(1+x³)¹⁰⁰ will have 100+1 = 101 terms

So (1+x)(1+x³)¹⁰⁰ will have 2×101 = 202 terms

Hence option c is the answer.

Question 10: The sum of the rational terms in the binomial expansion of (2^1/2 + 3^1/5)¹⁰ is

(a) 25

(b) 32

(c) 9

(d) 41

Solution:

(2^1/2 + 3^1/5)¹⁰ = ¹⁰C₀(2^1/2)¹⁰ + ¹⁰C₁(2^1/2)⁹(3^1/5) + … + ¹⁰C₁₀(3^1/5)¹⁰

Two rational terms are there. First and last term.

Sum of two rational terms = 2⁵ + 3²

= 32 + 9

= 41

Hence option d is the answer.

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JEE Binomial Theorem Previous Year Questions With Solutions