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

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

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

$=\left(1+\text{x}\right){\left(1+{\text{x}}^3\right)}^{100}$  $\left\{\because a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right\}$

${\left(1+{\text{x}}^3\right)}^{100}$ has $101$ terms

$\left(1+\text{x}\right){\left(1+{\text{x}}^3\right)}^{100}=1{\left(1+{\text{x}}^3\right)}^{100}+\text{x}{\left(1+{\text{x}}^3\right)}^{100}$

$\Rightarrow$ No. of  terms = $101+101=202$