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

Answer: 8.08

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