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

     $\because {(1+x)}^{101}{(1-x-x^2)}^{100}=(1+x){\left((1+x)(1-x+x^2)\right)}^{100}$      
 $=(1+x){(1+x^3)}^{100}$
 $=(1+x)(1{+}^{100}{C}_1{x}^3{+}^{100}{C}_2{x}^6{+}^{100}{C}_3{x}^9+\mathrm{...}+\mathrm{....}{+}^{100}{C}_{10}{x}^{300})$
Clearly, in this expression $x^3$ will present if  $n=3\lambda$ or  $n=3\lambda +1.$ So, n cannot be of the form $3\lambda +2$ .