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 $a_{0} + a_{2} + a_{4} + . . . + a_{2 n}$ .

Open in App

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,

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

Adding (1) and (2), we get
$3^{n} + 1 = 2 (a_{0} + a_{2} + . . . + a_{2 n})$

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

Suggest Corrections

Similar questions

(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{2 n} x^{2 n}

a_{0} + a_{2} + a_{4} + . . . . + a_{2 n}

(1 - x + x^{2})^{n} = a_{0} + a_{1} x + \dots + a_{2 n} x^{2 n}

a_{0} + a_{2} + a_{4} + \dots + a_{2 n}

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},

a_{0} + a_{1} + a_{2} + \dots . + a_{2 n}

a_{0} - a_{1} + a_{2} - a_{3} \dots . . + a_{2 n}

\frac{3^{n} + 1}{2}

\frac{3^{n} - 1}{2}

\frac{1 - 3^{n}}{2}

3^{n} + \frac{1}{2}

(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . . . . + a_{2 n} x^{2 n}

a_{0}, a_{1}, a_{2}, . . . . . . . . . . ., a_{2 n}

a_{n}

Join BYJU'S Learning Program