Prove for all $k \in \mathbb{N}$, that $\sum_{j=0}^{2k+1} {n+j-1\choose j} + \sum_{j=0}^{2k+1}(-1)^j{n+2k+2\choose j} = 0$

Question

Prove that this sum holds for all positive integers $k$. I'm quite sure this is right but I can't see immediately how to go about proving it. This will help resolve a problem regarding sums of binomial coefficients that I'm working on. Any ideas?

Answer 1

It's known that $\frac1{(1-x)^n}=\sum_{i\geq0}\binom{n+j-1}jx^j$. Combined with the relation $\frac1{1-x}\frac1{(1-x)^n}=\frac1{(1-x)^{n+1}}$, one finds that $\sum_{k\geq0}x^k\sum_{j=0}^k\binom{n+j-1}j=\sum_{k\geq0}x^k\binom{n+k}k$. In particular, one gathers that $$\sum_{j=0}^{2k+1}\binom{n+j-1}j=\binom{n+2k+1}{2k+1}. \tag1$$ On the other hand, the binomial relation $\binom{a}i+\binom{a}{i-1}=\binom{a+1}i$ and telescoping sums imply that \begin{align*} \sum_{j=0}^{2k+1}(-1)^j\binom{n+2k+2}j &=\sum_{j=0}^{2k+1}(-1)^j\left[\binom{n+2k+1}j+\binom{n+2k+1}{j-1}\right] \\ &=\sum_{j=0}^{2k+1}(-1)^j\binom{n+2k+1}j-\sum_{j=0}^{2k+1} (-1)^{j-1}\binom{n+2k+1}{j-1} \\ &=\sum_{j=0}^{2k+1}(-1)^j\binom{n+2k+1}j-\sum_{j=0}^{2k} (-1)^j\binom{n+2k+1}j\\ &=-\binom{n+2k+1}{2k+1}. \tag2 \end{align*} The OP's claim follows from adding (1) and (2).