Proving a sum of a strange series $ \sum_{i=1}^{n} 11i^{10}-55i^9+165i^8-330i^7+462i^6 -462i^5+330i^4-165i^3+55i^2-11i+1 = n^{11} $

Question

While messing around with zeta functions I encountered a strange sum: $$ \sum_{i=1}^{n} 11i^{10}-55i^9+165i^8-330i^7+462i^6 -462i^5+330i^4-165i^3+55i^2-11i+1 = n^{11} $$

How should I approach proving that this equality is true for all $n,i \in \mathbb{N}$ ? I ran this for a few values on some engines and it does work for every number I tried.

Answer 1

It's $$\sum_{i=1}^{n}(-(i-1)^{11}+i^{11})$$ and use the telescoping summation.

I used the following $$(x-1)^{11}=x^{11}-11x^{10}+55x^9-165x^8+330x^7-462x^6+462x^5-330x^4+165x^3-55x^2+11x-1$$ and $$\sum_{i=1}^{n}(-(i-1)^{11}+i^{11})=1^{11}-0^{11}+2^{11}-1^{11}+...+n^{11}-(n-1)^{11}=n^{11}.$$

Answer 2

factorial moments. As cumulative sums along a diagonal...

factorial moments.