Proof of the identity $ \sum_{i=n}^{N} \sum_{m=i}^{N} (-1)^{m-i} \binom{m}{i} x_m= \sum_{i=n}^{N} (-1)^{i-n} \binom{i-1}{n-1} x_i $

Question

I want to show the identity

$$ \sum_{i=n}^{N} \sum_{m=i}^{N} (-1)^{m-i} \binom{m}{i} x_m= \sum_{i=n}^{N} (-1)^{i-n} \binom{i-1}{n-1} x_i, $$

where $x_1, \ldots, x_n \in \mathbb{R}$.

By first changing the indices and then changing the order of summation, I got $$ \sum_{i=n}^{N} \sum_{m=i}^{N} (-1)^{m-i} \binom{m}{i} x_m= \sum_{m=1}^{N-n+1}x_m \sum_{i=1}^{m} (-1)^{m-i} \binom{m+n-1}{i+n-1}. $$ At this point I thought applying the Binomial theorem for $(1-1)^m$ could be useful, but I am not sure how to simplify the second sum.

Answer 1

Your change of indices doesn't look right, because you now have the term $x_1$ in your final sum. The interchange of summation should be: $$ \sum_{i=n}^N\sum_{m=i}^N=\sum_{m=n}^N\sum_{i=n}^m $$ Evaluating the RHS, we get $$ \begin{align} \sum_{m=n}^N\sum_{i=n}^m x_m(-1)^{m-i}\binom mi &=\sum_{m=n}^Nx_m(-1)^m\sum_{i=n}^m (-1)^i\binom mi\\ &=\sum_{m=n}^Nx_m(-1)^m\left[\sum_{i=0}^m (-1)^i\binom mi-\sum_{i=0}^{n-1} (-1)^i\binom mi\right]\\ &\stackrel{(1)}=\sum_{m=n}^Nx_m(-1)^m\left[0+(-1)^{n}\binom{m-1}{n-1}\right]\\ \end{align} $$ where (1) uses the binomial theorem on $(1-1)^m=0$ and the identity $$ \sum_{k=0}^s (-1)^k {n \choose k}=(-1)^s {n-1 \choose s}. $$ This should simplify to the desired result.