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7.Binomial Theorem
hard
$(1 +x)^{101} (1 +x^2 - x)^{100}$ ના વિસ્તરણમાં પદની સંખ્યા મેળવો.
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
