Binomial terms count

Using the well known identity , $x^3+1=(x+1)(x^2-x+1)$ , $\begin{aligned}(x+1)^{101}(x^2-x+1)^{100} &=(x^3+1)^{100}(x+1) \\ &= x(x^3+1)^{100}+(x^3+1)^{100} \end{aligned}$ From the binomial theorem , we have, $\; \;\;\;\;\;\;\;\;\;\;(x^3+1)^{100}=\binom{100}{0}x^0+\binom{100}{1}x^3+\binom{100}{2}x^6+\cdots+\binom{100}{99}x^{297}+\binom{100}{100}x^{300}$ $\implies x(x^3+1)^{100}=\binom{100}{0}x^1+\binom{100}{1}x^4+\binom{100}{2}x^7+\cdots+\binom{100}{99}x^{298}+\binom{100}{100}x^{301}$ Observe that $(x^3+1)^{100}$ and $x(x^3+1)^{100}$ have no common powers of $x$ in their expansion, so their sum cannot be simplified any further (the terms cannot be combined). Both of the above expansions have $101$ terms. Therefore, the total number of terms is $\boxed{202}$