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

30.00

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30.0

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Solution

Let $(1 + x + x^{2} + . . . + x^{2 n}) (1 - x + x^{2} - . . . + x^{2 n}) = a_{0} + a_{1} x + a_{2} x^{2} + . . .$
Put $x = 1$
$2 n + 1 = a_{0} + a_{1} + a_{2} + a_{3} + . . . (1)$

Put $x = - 1$
$2 n + 1 = a_{0} - a_{1} + a_{2} - a_{3} + . . . (2)$
Adding (1) and (2)
$2 (2 n + 1) = 2 (a_{0} + a_{2} + a_{4} + . . .$
$\Rightarrow 2 n + 1 = 61$
$\Rightarrow n = 30$

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