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?



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


