Prove that $\sum\_{k=0}^n {k+1\choose 2}^R + \sum\_{k=0}^{-n-2} {k+1\choose 2}^R = 0.$ This can be shown using Faulhaber's formula but it's very long. Is there a nicer, shorter method? Any thoughts or ideas?



Prove that $\sum_{k=0}^n {k+1\choose 2}^R + \sum_{k=0}^{-n-2} {k+1\choose 2}^R = 0.$
The "correct" definition of $\sum\_{k=0}^{-n-2} f(k)$ is $-\sum\_{k=-n-1}^{-1}f(k)$. Now ${-a+1\choose 2}={a\choose 2}$. Thus the terms of $\sum\_{k=0}^{-n-2}{k+1\choose 2}^R$ cancel with the terms of $\sum\_{k=0}^n{k+1\choose 2}^R$.


