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

English

Gujarati

Hindi

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

Share

Similar Questions

If the number of terms in the expansion of ${\left( {1 – \frac{2}{x} + \frac{4}{{{x^2}}}} \right)^n},x \ne 0$ is $28$ then the sum of the coefficients of all the terms in this expansion, is :

hard

(JEE MAIN-2016)

$\frac{1}{{1!(n – 1)\,!}} + \frac{1}{{3!(n – 3)!}} + \frac{1}{{5!(n – 5)!}} + …. = $

medium

In the expansion of

$(2x + 1).(2x + 5) . (2x + 9) . (2x + 13)…(2x + 49),$ find the coefficient of $x^{12}$ is :-

normal

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + … + {C_n}{x^n}$, then the value of ${C_0} + {C_2} + {C_4} + {C_6} + …..$ is

easy

If the sum of the coefficients in the expansion of ${(\alpha {x^2} – 2x + 1)^{35}}$ is equal to the sum of the coefficients in the expansion of ${(x – \alpha y)^{35}}$, then $\alpha $=

hard