First slide

Binomial theorem for positive integral Index

Question

if ${(1 - x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$ then $a_{0} + a_{2} + a_{4} + \dots + a_{2 n}$ is equal to

Easy

A

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

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

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

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

Solution

${(1 - x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$ on putting X=1, we get
$(1 - 1 + 1)^{n} = a_{0} + a_{1} + a_{2} + \dots + a_{2 n}$
$\Rightarrow 1 = a_{0} + a_{1} + a_{2} + \dots + a_{2 n} . . . . . (i)$
Again, putting x= - 1, we get
$3^{n} = a_{0} - a_{1} + a_{2} - \dots + a_{2 n}$
$\frac{3^{n} + 1}{2} = a_{0} + a_{2} + a_{4} + \dots + a_{2 n}$

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