Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately

Question

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.

Answer 1

Given that $(-1)^k=1$ for even $k$, and $-1$ for odd $k$, I'd suggest splitting your sum into $$\sum_{n=0}^{10}{\binom{2n+2}{2}}-\sum_{n=1}^{10}{\binom{2n+1}{2}}$$ with the former representing even $k$ and the latter for odd $k$.

Answer 2

There is! You can use the identity $\binom{k+2}2 = \sum_{i=0}^{k+1}i$.

Our sum is $$\sum_{k=0}^n (-1)^k \binom{k+2}{2} = \sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i$$

For odd $n$: Letting $m=\frac {n-1}2$ and pairing up the terms we get

$$\begin{align}\sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i &= \sum_{j=0}^m [(-1)^{2j}\sum_{i=0}^{2j+1}i + (-1)^{2j+1}\sum_{i=0}^{2j+2}i] \\&=\sum_{j=0}^m [\sum_{i=0}^{2j+1}i -\sum_{i=0}^{2j+2}i] \\&= \sum_{j=0}^m-(2j+2) \\&= -2\sum_{j=0}^m(j+1) \\&= -2\sum_{j=1}^{m+1}j\\ &= -2[\binom{m+2}2-0] \\&= -(m+2)(m+1) \\&=-\frac{(n+3)(n+1)}4 \end{align}$$

For even $n$: $$\begin{align} \sum_{k=0}^n (-1)^k \sum_{i=0}^{k+1}i &= \sum_{k=0}^{n-1} (-1)^k \sum_{i=0}^{k+1}i + \sum_{i=0}^{n+1}i \\&= -\frac{((n-1)+3)((n-1)+1)}4 + \binom{n+2}2 \\ &=-\frac{n^2+2n}4 + \frac{n^2 + 3n + 2}{2} \\&= \frac{ - n^2 - 2n + 2n^2 + 6n + 4}4 \\&=(\frac{n+2}2)^2 \end{align}$$

Answer 3

Alternatively: $$\begin{align}\sum_{k=0}^{20}(-1)^k\binom{k+2}2=&\sum_{k=0}^{20}\binom{k+2}2-2\cdot \sum_{k=0}^{9}\binom{2k+3}2=\\ &{23\choose 3}-\sum_{k=0}^9 (2k+3)(k+1)=\\ &1771-2\cdot 825=121.\end{align}$$

Answer 4

Using the identity from Pascal's Triangle, we get $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{k+a}{b} &=\sum_{k=0}^n(-1)^k\left[\binom{k+a+1}{b}-\binom{k+a}{b-1}\right]\\ &=\sum_{k=1}^{n+1}(-1)^{k-1}\left[\binom{k+a}{b}-\binom{k+a-1}{b-1}\right]\\ &=\frac12\left(\binom{a}{b}+(-1)^n\binom{n+a+1}{b}-\sum_{k=1}^{n+1}(-1)^{k-1}\binom{k+a-1}{b-1}\right)\\ &=\frac12\left(\binom{a}{b}+(-1)^n\binom{n+a+1}{b}-\sum_{k=0}^n(-1)^k\binom{k+a}{b-1}\right)\tag1 \end{align} $$ Multiplying by $(-2)^b$, we get $$ \begin{align} (-2)^b\sum_{k=0}^n(-1)^k\binom{k+a}{b} &=\frac12(-2)^b\left[\binom{a}{b}+(-1)^n\binom{n+a+1}{b}\right]\\ &+(-2)^{b-1}\sum_{k=0}^n(-1)^k\binom{k+a}{b-1}\\ &=\frac12\sum_{j=1}^b(-2)^j\binom{a}{j}+\frac{(-1)^n}2\sum_{j=1}^b(-2)^j\binom{n+a+1}{j}\\ &+[n\equiv0\bmod2]\\ &=\bbox[5px,border:2px solid #C0A000]{\frac12\sum_{j=0}^b(-2)^j\binom{a}{j}+\frac{(-1)^n}2\sum_{j=0}^b(-2)^j\binom{n+a+1}{j}}\tag2 \end{align} $$ Setting $a=b=2$ yields $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{k+2}{2} &=\frac18\left[\binom{2}{0}-2\binom{2}{1}+4\binom{2}{2}\right]\\ &+\frac{(-1)^n}8\left[\binom{n+3}{0}-2\binom{n+3}{1}+4\binom{n+3}{2}\right]\\[6pt] &=\frac18+\frac{(-1)^n}8\left(2n^2+8n+7\right)\tag3 \end{align} $$ Setting $n=20$ yields $$ \begin{align} \sum_{k=0}^{20}(-1)^k\binom{k+2}{2} &=\frac18+\frac{(-1)^{20}}8\left(2\cdot20^2+8\cdot20+7\right)\\[6pt] &=121\tag4 \end{align} $$

Answer 5

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

$$ = 1 + 3 + 5 + \ldots + 21 = 121$$