First slide

Binomial theorem for positive integral Index

Question

Sum of coefficients of terms of odd powers of $x$ in ${(1 + x - x^{2} - x^{3})}^{8}$ is

Easy

A

0
B

1
C

2
D

–1

Solution

Given expansion ${(1 + x - x^{2} - x^{3})}^{8}$ is

Let consider $f (x) = {(1 + x - x^{2} - x^{3})}^{8}$ substitute $x = 1$ then we get

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

Similarly let consider $f (x) = {(1 + x - x^{2} - x^{3})}^{8}$ substitute $x = - 1$ then we get

$\Rightarrow f (- 1) = {(1 - 1 - 1 + 1)}^{8} = 0$

Sum of coefficients of odd powers of $x = \frac{f (1) - f (- 1)}{2} = \frac{0 - 0}{2}$

                                                                            $= 0$

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