If the sum of the coefficients of all even powers of x in the product (1 + x + x ^ 2 + + x ^ 2n ) (1 - x + x ^ 2 - x ^ 3 + + x ^ 2n ) is 61, then n is equal to

Correct answer: 0 Let (1 - x + x ^ 2 . .)(1 + x + x ^ 2 ) = a _ 0 + a _ 1 × a _ 2 x ^ 2 + Put x = 1 , 1(2n + 1) = a _ 0 + a _ 1 + a _ 2 + . . a _ 2n (2) Put x = - 1 (2n + 1) × 1 = a _ 0 - a _ 1 + a _ 2 + . . a _ 2n (2) From (1) + (2) 4n + 2 = 2 ( a _ 0 + a _ 2 + ) = 2 × 61 ⇒ 2n + 1 = 61 ⇒ n = 30

JEE Main 20207 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2020 · 7 January · Morning Shift · Mathematics · Binomial Theorem

If the sum of the coefficients of all even powers of x in the...

If the sum of the coefficients of all even powers of

x

in the product \left(1 + x + {x}^{2} + \ldots + {x}^{2n}\right) \left(1 - x + {x}^{2} - {x}^{3} + \ldots + {x}^{2n}\right) is

61,

then n is equal to

Correct answer

Step-by-step solution

Let

\left(1 - x + {x}^{2} . .\right)\left(1 + x + {x}^{2} \ldots\right) = {a}_{0} + {a}_{1} \times {a}_{2}{x}^{2} + \ldots

Put

x = 1

1\left(2n + 1\right) = {a}_{0} + {a}_{1} + {a}_{2} + . . {a}_{2n} \ldots \left(2\right)

Put

x = - 1

\left(2n + 1\right) \times 1 = {a}_{0} - {a}_{1} + {a}_{2} + . . {a}_{2n} \ldots \left(2\right)
From \left(1\right) + \left(2\right)

4n + 2 = 2 \left({a}_{0} + {a}_{2} + \ldots\right)

= 2 \times 61

⇒ 2n + 1 = 61 ⇒ n = 30

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