The number of terms in the expansion of (1+x)^101(1+x^2-x)^100 in powers of x is:

Question

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

Accepted Answer

$$(1+x)^{101}(1-x+x^2)^{100}=(1+x)(1+x)^{100}(1-x+x^2)^{100}=(1+x)\left[(1+x)(1-x+x^2)\right]^{100}=(1+x)(1+x^3)^{100}$$

$(1+x^3)^{100}$ has 101 terms; multiplying by $(1+x)$ doubles the term count (no overlaps since powers differ by 1 mod 3 vs 0 mod 3):

$$2\times101=202\text{ terms}$$

Solutions · 1