Jugular Expansion

Relevant wiki: Properties of Binomial Coefficients

Write $\small{(1+x)^{101}=(1+x)(1+x)^{100}}$ so that: $\large (1+x)\left[(1+x)(1+x^2-x)\right]^{100}$ $\large =(1+x)\left[1+x^3\right]^{100}$ $\large =\left[1+x^3\right]^{\color{#D61F06}{100}}+x\left[1+x^3\right]^{\color{#D61F06}{100}}$

(The first bracket will give $101$ independent terms of the form $x^{3n}$ which cannot be grouped with any of the term of second bracket since they will be of the form $x^{3n+1}$ $~~(n\in\mathbb{Z^+}\cup \{0\})$ .)

Thus we would get a total of $\large 101+101=\boxed{\color{#20A900}{202}}$ terms.