If coefficient of $x^n$ in $(1+x{)}^{101}{\left(1-x+x^2\right)}^{100}$ is non-zero, then n cannot be of the form

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

$\begin{array}{l}=(1+x){\left((1+x)\left(1-x+x^2\right)\right)}^{100}\\ =(1+x){\left(1+x^3\right)}^{100}\\ =(1+x)\left\{C_0+C_1{x}^3+C_2{x}^6+\dots +C_{100}{x}^{300}\right\}\\ =(1+x)\sum _{r=0}^{100} n_r{x}^{3r}=\sum _{r=0}^{100}{ }^n{C}_r{x}^{3r}+\sum _{r=0}^{100}{ }^n{C}_r{x}^{3r+1}\end{array}$

Hence there will be no term containing $3r+2$