Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20}$ in $(1+x^2+x^4\dots)(1+x+x^2\dots)^2$ which simplified to the term above.
I know $\sum_{k=0}^{20}\binom{k+2}{2}=\binom{23}3$ but the $(-1)^k$ is ruining things.

$$\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2 = \binom{2}{2} \underbrace{-\binom{3}{2} + \binom{4}{2}}_{\binom{3}{1}} \underbrace{-\binom{5}{2} + \binom{6}{2}}_{\binom{5}{1}}- \ldots \underbrace{-\binom{21}{2} + \binom{22}{2}}_{\binom{21}{1}}$$
$$ = 1 + 3 + 5 + \ldots + 21 = 121$$
