JEE Main Binomial Theorem and Mathematical Induction Previous Year Questions

Binomial Theorem and Mathematical Induction are two important concepts in JEE preparation. In algebra, the Binomial Theorem focuses on the expansion of powers on any binomial expression. The Binomial theorem describes the algebraic expansion of powers of a binomial whereas Mathematical Induction is a technique which is used to prove a mathematical formula, a statement or a theorem is true for all the natural numbers. This mathematical technique basically includes three steps to prove a statement. This article covers the Binomial theorem, the formula for the general term in binomial expansion, the formula for the middle term in binomial expansion and the Binomial formula of some expansions. The Binomial Theorem and Mathematical Induction questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. These questions include all the important topics and formulae. About 1-2 questions are asked from this chapter in the examination every time.

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Binomial Theorem and Mathematical Induction Previous Year Questions with Solutions

Question 1: If (1 + ax)ⁿ = 1 + 8x + 24x² +…., then the value of a and n are ________.

Solution:

As given (1 + ax)ⁿ = 1 + 8x + 24x² +….

On comparing same coefficients, we have

⇒ na (n − 1) a = 48

8 (8 − a) = 48

8 − a = 6

⇒ a = 2

⇒ n = 4

Question 2: If the coefficient of x⁷ in (ax² + [1/bx])¹¹ is equal to the coefficient of x⁻⁷ in (ax − 1/bx²)¹¹, then ab = __________.

Solution:

In the expansion of (ax² + [1/bx])¹¹, the general term is

T_{r + 1} = ¹¹C_r (ax²)^{11 − r} (1/bx)^r

= ¹¹C_r * a^{11 − r} * [ 1/b^r ] * x^{22 − 3r}

For x⁷, we must have 22 – 3r = 7

r = 5, and the coefficient of x⁷ = ¹¹C₅ * a¹¹⁻⁵ * [1/b⁵] = ¹¹C₅ * a⁶ * b⁵

Similarly, in the expansion of (ax⁻¹bx²)¹¹, the general term is

T_{r + 1} = ¹¹C_r * (−1)^r * [a^{11 − r}/b^r ] * x^{11 − 3r}

For x^-7 we must have, 11 – 3r = -7 , r = 6, and the coefficient of x⁻⁷ is ¹¹C₆ * [a⁵/b⁶] = ¹¹C₅ * a₅ * b₆.

As given, ¹¹C₅ [a⁶/b⁵] = ¹¹C₅ * [a⁵/b⁶]

⇒ ab = 1

Question 3: Find the value of

Solution:

(1 − x)³⁰ = ³⁰C₀x₀ − ³⁰C₁x₁ + ³⁰C₂x² + . . . . . . + (−1)^{30 30}C₃₀x³⁰ ….(i)

(x + 1)³⁰ = ³⁰C₀ x³⁰ + ³⁰C₁x²⁹ + ³⁰C₂x²⁸ + . . . . . . + ³⁰C₁₀x²⁰ + . . . . + ³⁰C₃₀x⁰ ….(ii)

Multiplying (i) and (ii) and equating the coefficient of x²⁰ on both sides, we get

required sum = coefficient of x²⁰ in (1 – x²)³⁰ = ³⁰C₁₀.

Question 4: If

Solution:

Question 5: In the polynomial (x − 1) (x − 2) (x − 3) . . . . . (x − 100), what is the coefficient of x⁹⁹?

Solution:

(x − 1) (x − 2) (x − 3) . . . . . (x − 100)

Number of terms = 100

Coefficient of x⁹⁹ = (x − 1) (x − 2) (x − 3) . . . . . (x − 100)

= (−1 −2 −3 −. . . . . . −100)

= − (1 + 2 + . . . . . +100)

= − [100 * 101 / 2

= – 5050

Question 6: In the expansion of (x + a)ⁿ, the sum of odd terms is P and sum of even terms is Q, then the value of (P² − Q²) will be _________.

Solution:

(x + a)ⁿ = xⁿ + ⁿC₁xⁿ⁻¹ a + . . . . . = (xⁿ + ⁿC₂xⁿ⁻²a² + . . . . . . . + (ⁿC₁xⁿ⁻¹a + ⁿC₃xⁿ⁻³a³ + . . . . .) = P+Q

(x − a)ⁿ = P − Q

As the terms are altered,

P² − Q² = (P + Q) (P − Q) = (x + a)ⁿ (x − a)ⁿ

P² − Q² = (x² − a²)ⁿ

Question 7: If the sum of the coefficients in the expansion of (1 − 3x + 10x²)ⁿ is a and if the sum of the coefficients in the expansion of (1 +x²)ⁿ is b, then what is the relation between a and b?

Solution:

We have a = sum of the coefficient in the expansion of (1 − 3x + 10x²)ⁿ = (1 − 3 + 10)ⁿ = (8)ⁿ

(1 − 3x + 10x²)ⁿ = (2)³ⁿ, [Putting x = 1] .

Now, b = sum of the coefficients in the expansion of (1 + x²)ⁿ = (1 + 1)ⁿ = 2n.

Clearly, a = b³

Question 8: (1 + x)ⁿ − nx − 1 is divisible by (where n∈N)

A) 2x

B) x²

C) 2x³

D) All of these

Solution:

(1 + x)ⁿ = 1 + nx + ([n (n − 1)] / [2!]) * x² + ([n (n − 1) (n − 2)] / [3!]) * x³ + . . . . .

(1 + x)ⁿ − nx − 1 = x² [([n (n − 1)] / [2!]) + ([n (n − 1) (n − 3)] / [3!]) * x + . . . . .]

From above it is clear that (1 + x)ⁿ − nx − 1 is divisible by x².

Trick: (1 + x)ⁿ − nx − 1, put n = 2 and x = 3;

Then 4² − 2 * 3 − 1 = 9 is not divisible by 6, 54 but divisible by 9, which is given by option (b) i.e., x² = 9.

Question 9: If the three consecutive coefficients in the expansion of (1 + x)ⁿ are 28, 56 and 70, then the value of n is ______.

Solution:

Let the three consecutive coefficients be ⁿC_r−1 = 28, ⁿC_r = 56 and ⁿC_r+1 = 70, so that

ⁿC_r / ⁿC_r−1 = [n − r + 1] / [r] = 56 / 28 = 2 and ⁿC_r+1 / ⁿC_r = [n − r] / [r + 1] = 70 / 56 = 5 / 4

This gives n + 1 = 3r and 4n − 5 = 9r

4n − 5 / n + 1 = 3

⇒ n = 8

Question 10: Let R = (5√5 + 11)^{2n + 1} and f = R − [R], where [.] denotes the greatest integer function. The value of R * f is

Solution:

Since (5√5 − 11) (5√5 + 11) = 4

5√5 − 11 = 4 / (5√5 + 11), Because 0 < 5√5 − 11 < 1 ⇒ 0 < (5√5 − 11)^{2n + 1} < 1, for positive integral n.

Again, (5√5 + 11)^{2n + 1} − (5√5 − 11)^{2n + 1} = 2 {²ⁿ⁺¹C₁ (5√5)²ⁿ * 11 + ^{2n + 1}C₃(5√5)^{2n − 2} × 11³ + . . .. +

^{2n + 1}C_2n+1 11²ⁿ⁺¹}

= 2 {²ⁿ⁺¹C₁(125)ⁿ * 11 + ²ⁿ⁺¹C₃(125)ⁿ⁻¹ 11³ +. . . . . . + ²ⁿ⁺¹C_2n+1 11²ⁿ⁺¹}

= 2k, (for some positive integer k)

Let f′= (5√5 − 11)^{2n + 1} , then [R] + f − f′=2k

f − f′ = 2k − [R]

⇒ f − f′ is an integer.

But, 0 ≤ f < 1; 0 < f′<1

⇒ −1 < f − f′ < 1

f − f′ = 0 (integer)

f = f′

Therefore, Rf = Rf′ = (5√5 + 11)^{2n + 1} * (5√5 − 11)^{2n + 1}

= ([5√5]² + 11²)^{2n + 1}

= 4²ⁿ⁺¹

Question 11: The digit in the unit place of the number (183!) + 3¹⁸³ is _____.

Solution:

We know that n! terminates in 0 for n³ at 5 and 3⁴ⁿ terminator in 1, (because 3⁴ = 81)

Therefore, 3¹⁸⁰ = (3⁴)⁴⁵ terminates in 1.

Also 3³ = 27 terminates in 7

3¹⁸³ = 3¹⁸⁰ * 3³ terminates in 7.

183! + 3¹⁸³ terminates in 7 i.e. the digit in the unit place = 7.

Question 12: If the coefficients of p^th, (p + 1)^th and (p + 2)^th terms in the expansion of (1 + x )ⁿ are in A.P., then find the equation in terms of n.

Solution:

Coefficient of p^th, (p + 1)^th and (p + 2)^th terms in expansion of (1 + x )ⁿ are ⁿC_p−1, ⁿC_p ,ⁿC_p+1. Then 2ⁿC_p = ⁿC_p−1 + ⁿC_p+1

⇒n² − n (4p + 1) + 4p² − 2 = 0

Trick: Let p = 1, hence ⁿC₀, ⁿC₁ and ⁿC₂ are in A.P.

⇒ 2 * ⁿC₁ = ⁿC₀ + ⁿC₂

⇒ 2n = 1 + [n (n − 1)] / [2]

⇒ 4n = 2 + n² − n

⇒ n² − 5n + 2 = 0

Question 13: The interval in which x must lie so that the greatest term in the expansion of (1 + x)²ⁿ has the greatest coefficient, is _____.

Solution:

Here the greatest coefficient is ²ⁿC_n.

Therefore, ²ⁿC_nxⁿ > ²ⁿC_n+1xⁿ⁻¹ ⇒ x > n / [n + 1] and ²ⁿC_nxⁿ > ²ⁿC_n-1xⁿ⁺¹

⇒ x < [n + 1] / n

Hence, the required interval is (n / [n + 1], [n + 1] / n).

Question 14: The approximate value of (1.0002) 3000 is __________.

Solution:

(1.0002) 3000 = (1 + 0.0002) 3000

= 1 + (3000) (0.0002) + ([(3000) (2999)] / [1.2]) * (0.0002)²

= 1 + (3000) (0.0002)

= 1.6

Question 15: The last digit in 7³⁰⁰ is __________.

Solution:

We have 7² = 49 = 50 − 1

Now, 7³⁰⁰ = (7²)¹⁵⁰

= (50 − 1)¹⁵⁰

= ¹⁵⁰C₀ (50)¹⁵⁰ (−1)⁰ + ¹⁵⁰C₁ (50)¹⁴⁹ (−1)¹ + . . . . . .+ ¹⁵⁰C₁₅₀ (50)⁰ (−1)¹⁵⁰

Thus, the last digits of 7³⁰⁰ are ¹⁵⁰C₁₅₀ * 1 * 1 i.e., 1.

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Binomial theorem – Solved Questions

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