# Simplify $\sum_{i=0}^n(-1)^i { n \choose i } (n - i)^3$

How can we simplify $$\sum_{i=0}^n(-1)^i { n \choose i } (n - i)^3$$?

I started computing the first values to figure the pattern:

$$\sum_{i=0}^0(-1)^i { 0 \choose i } (0 - i)^3 = 0$$

$$\sum_{i=0}^1(-1)^i { 1 \choose i } (1 - i)^3 = 1 + 0 = 1$$

$$\sum_{i=0}^2(-1)^i { 2 \choose i } (2 - i)^3 = 2^3 - 2 + 0 = 6$$

$$\sum_{i=0}^3(-1)^i { 3 \choose i } (3 - i)^3 = 3^3 -3 \cdot 2^3+3-0 = 6$$

The results don't lead to a pattern and so I thought I must be missing something.

I know that $$\sum_{i=0}^n(-1)^i { n \choose i } = 0$$ but this also doesn't help.

How to go forth here to solve this (finding a form without the $$\Sigma$$)?

The result is related to the Stirling number of the second kind: $$n!{3 \brace n}.$$ Particularly for $$n>3$$ all values are $$0$$.

I will re-index the sum for personal taste.

The trick here, from the theory of generating functions, is to write $$k^3$$ as the linear differential operator $$(x\frac{\partial}{\partial x})$$ applied thrice to the power $$x^k$$, then evaluated at $$x=1$$. Then

$$\sum_{k=0}^n (-1)^k \binom{n}{k}k^3=\left(x\frac{\partial}{\partial x}\right)^3\left.\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}x^k\right|_{x=1}$$

$$= \left.\left(x\frac{\partial}{\partial x}\right)^3 (x-1)^n\right|_{x=1}$$

Funnily enough, if $$n>3$$ then all of the derivatives of $$(x-1)^n$$ above (using the product rule multiple times) will still retain a $$(x-1)$$ factor, and so when evaluated at $$x=1$$ will yield $$\Sigma=0$$. You will see this pattern if you evaluate the sum for $$n=4,5,6,\cdots$$ (you will get $$0$$ each time)!

First note that, for $$n\ge 3$$, $${ n \choose i }i(i-1)(i-2)=n(n-1)(n-2){ n-3 \choose i-3 }.$$

So, for $$n\ge 4$$, $$\sum_{i=0}^n(-1)^i { n \choose i } i(i-1)(i-2)=-n(n-1)(n-2)\sum_{j=0}^{n-3}(-1)^j { {n-3} \choose{j} }=0.$$ Similarly $$\sum_{i=0}^n(-1)^i { n \choose i } i(i-1)=\sum_{i=0}^n(-1)^i { n \choose i } i=\sum_{i=0}^n(-1)^i { n \choose i }=0.$$ Therefore, by expanding $$(n-i)^3$$, $$\sum_{i=0}^n(-1)^i { n \choose i } (n-i)^3=0.$$