First slide

Binomial theorem for positive integral Index

Question

Sum of coefficients of terms of even powers of x in ${(1 + x + x^{2} + x^{3})}^{5}$ is

Easy

A

512
B

516
C

612
D

234

Solution

Given expansion ${(1 + x + x^{2} + x^{3})}^{5}$ is

Sum of coefficients of terms of even powers of x in ${(1 + x + x^{2} + x^{3})}^{5}$ is

Let consider $f (x) = {(1 + x + x^{2} + x^{3})}^{5}$ substitute $x = 1$ then we get

$\Rightarrow f (1) = {(1 + 1 + 1 + 1)}^{5} =$ ${(4)}^{5}$

Similarly let consider $f (x) = {(1 + x + x^{2} + x^{3})}^{5}$ substitute $x = - 1$ then we get

$\Rightarrow f (- 1) = {(1 - 1 + 1 - 1)}^{5} = 0$

Sum of coefficients of even powers of $x$ $= \frac{f (1) + f (- 1)}{2}$

$= \frac{4^{5} + 0}{2} = 2^{9} = 512$

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