Number of terms in expansion of (1 +x)^{101} (1 +x^2 - x)^{100} in powers of x is

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7.Binomial Theorem

hard

The number of terms in the expansion of $(1 +x)^{101} (1 +x^2 - x)^{100}$ in powers of $x$ is

A

$302$

B

$301$

C

$202$

D

$101$

(JEE MAIN-2014)

Solution

Given expansion is $(1+x)^{101}\left(1-x+x^{2}\right)^{100}$

$=(1+x)(1+x)^{100}\left(1-x+x^{2}\right)^{100}$

$=(1+x)\left[(1+x)\left(1-x+x^{2}\right)\right]^{100}$

$=(1+x)\left[\left(1-x^{3}\right)^{100}\right]$

Expansion $\left(1-x^{3}\right)^{100}$ will have

$100+1$ $=101$ terms

$\mathrm{So},(1+x)\left(1-x^{3}\right)^{100}$ will have

$2 \times 101$

$=202$ terms

Standard 11

Mathematics

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