If $(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{2 n} x^{2 n}$ where $a_{0}, a_{1}, a_{2}, . . ., a_{2 n}$ are in A.P. then prove that $a_{n} = \frac{1}{2 n + 1}$

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Solution

$a_{n} = T_{n + 1}$ of A.P. = $a_{0} + n d$
Putting x = 1 in the given relation, we have
$a_{0} + a_{1} + a_{2} + . . . . . a_{2 n} = 1^{n}$
or $S_{2 n + 1} = \frac{2 n + 1}{2} [a_{0} + a_{2 n}] = 1$
or $a_{0} + T_{2 n + 1} = \frac{2}{2 n + 1} ∵ a_{2 n} = T_{2 n + 1}$
or $a_{0} + (a_{0} + 2 n d) = \frac{2}{2 n + 1}$
or $2 (a_{0} + 2 n d) = \frac{2}{2 n + 1} o r a_{0} + n d = \frac{1}{2 n + 1}$
or $T_{n + 1} = a_{0} + n d = a_{n} = \frac{1}{2 n + 1}$

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