Let $[x]$ denote greatest integer less than or equal to $x$. If for $n \in N,\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\sum_{\mathrm{j}=0}^{\left[\frac{3 n}{2}\right]} \mathrm{a}_{2 \mathrm{j}}+4 \sum_{\mathrm{j}=0}^{\left[\frac{3 n-1}{2}\right]} \mathrm{a}_{2 \mathrm{j}+1}$ is equal to :

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Binomial Theorem

Let

[x]

denote greatest integer less than or equal to

x

. If for

n \in N, (1 - x + x^{3})^{n} = \sum_{j = 0}^{3 n} a_{j} x^{j}

, then

\sum_{j = 0}^{[\frac{3 n}{2}]} a_{2 j} + 4 \sum_{j = 0}^{[\frac{3 n - 1}{2}]} a_{2 j + 1}

is equal to :

(A)1✓ Correct
(B)n
(C) $2^{n - 1}$
(D)2

Answer: (A)

Solution

Putting

x = 1

and

x = - 1

(1 - x + x^{3})^{n} = \sum_{j = 0}^{3 n} a_{j} x^{j}

j = 0 \sum 3 n a_{j} = (1 - 1 + 1)^{n} = 1, j = 0 \sum 3 n (- 1)^{j} a_{j} = (1 + 1 - 1)^{n} = 1

The largest even index not exceeding

3 n

2 [\frac{3 n}{2}]

and the largest odd one is

2 [\frac{3 n - 1}{2}] + 1

, so the two sums in the expression are precisely the sum of the even indexed coefficients and the sum of the odd indexed ones. Adding and subtracting the two relations,

\sum a_{2 j} = 1, \sum a_{2 j + 1} = 0

so the expression equals

1 + 4 \cdot 0 = 1

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Binomial Theorem

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