Question

If the number of terms in the expansion of $\left(1 + x\right)^{101}\left(1 + x^{2} - x\right)^{100}$ is $n$ , then the value of $\frac{n}{25}$ is equal to

Answer 1

$\left(1 + x\right)^{101}\left(1 + x^{2} - x\right)^{100}$
$=\left(1 + x\right)\left(\left(1 + x\right) \left(1 - x + x^{2}\right)\right)^{100}$
$=\left(1 + x\right)\left(1 + x^{3}\right)^{100}$
$=1\times \left(1 + x^{3}\right)^{100}+x\times \left(1 + x^{3}\right)^{100}$
So, the number of terms $=$ ( $101$ terms of the form $x^{3 k}$ ) + ( $101$ terms of the form $x^{3 k + 1}$ )
$=202$ terms
$\Rightarrow n=202$