If $(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . a_{2 n} x^{2 n}$ , find the value of $a_{0} + a_{2} + a_{4} + . . . . . + a_{2 n} .$

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Solution

Putting x =1 and -1 in $(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{2 n} x^{2 n}$ we get,

$1 = a_{0} + a_{1} + a_{2} + . . . + a_{2 n} . . . . . (i)$

and

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

Adding (i) and (ii), we get

$3^{n} + 1 = 2 (a_{0} + a_{2} + . . . . + a_{2 n})$

Hence, the values of $a_{0} + a_{2} = a_{4} + . . . . a_{2 n}$ is $\frac{3^{n} + 1}{2}$

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