$\sum_{k=0}^n k(k − 1)(k − 2)=1/4(n+1)n(n-1)*(n-2)$
Find a general formula for $\sum_{k=0}^n k(k − 1)(k − 2) · · · (k − r + 1)$ with respect to k and r.
I know summation of power series such as $\sum_{k=0}^n k^2=(n+1)n(2n+1)/6$, is there any way i can use the power series formula to find a general formula for $\sum_{k=0}^n k(k − 1)(k − 2) · · · (k − r + 1)$?
Surprisingly, yes.
Your expression is $\displaystyle \sum_{k=0}^n k(k-1)(k-2)...(k-r+1)$ which is equivalent to $\displaystyle \sum_{k=0}^n \frac{k!}{(k-r)!}$, which is equal to $\displaystyle r!\sum_{k=0}^n \frac{k!}{r!(k-r)!}=\displaystyle r!\sum_{k=0}^n \binom{k}{r}$.
This term $\displaystyle \sum_{k=0}^n \binom{k}{r}$ is known as the Hockey-stick identity***, and is equal to $\displaystyle \binom{n+1}{r+1}$.
So your final answer is $\displaystyle \boxed{r!\binom{n+1}{r+1}}$.
** The Hockey stick identity actually has a k=r, not a k=0 in the bottom of the sum. however, if k < r, k choose r = 0, so we don't have an issue with this notation.