Formula for the general Cavalieri Sum: $S_n(p)=\sum\limits_{k=1}^{n} k^p\,\,\,n, p\in\mathbb N$ What kind of formula is there that can be used for calculating the sum of power of $x$ of numbers from $1$ to $a$?
I know that the sum of numbers from $1$ to $a$ is $\ (n^2 + n)/ 2  \ $ and that the sum of the squared numbers from $1$  to $a$ is $n^3/3+n^2/2+n/6 $.
If possible, please give a solution that excludes Bernoulli numbers and works for negative numbers too.
 A: As far as I am aware of, there is no general formula.
There are several methods to compute these sums however, i'll mention my favorite one:
You find a polynomial $P$ of degree $n+1$, such that $P(x+1)-P(x)=x^n$. This basically means solving a system of linear equations and one can prove using elementary linear algebra, that such a polynomial always exists.
Once you have it, then you simply use telescopic sums: $\sum \limits_{k=0}^a k^n=\sum \limits_{k=0}^a (P(k+1)-P(k))=P(a+1)-P(0)$.
A: Let $$ S_n(p)=\sum_{k=1}^{n} k^p\qquad n, p\in\mathbb N ~~~~~\text{called Cavalieri sum of oder p}$$
then, 
We know the following Binomial formula 
$$ (k+1)^p = k^p+ \sum_{i=0}^{p-1}\binom{p}{i} k^i$$
where $\binom{p}{i}= \frac{p!}{i!(p-i)!}$.
Which implies that, 
$$\sum_{k=1}^{n} (k+1)^p =\sum_{k=1}^{n} k^p+\sum_{i=0}^{p-1}\binom{p}{i} \sum_{k=1}^{n} k^i = S_n(p) +\sum_{i=0}^{p-1}\binom{p}{i} S_n(i) $$
But $$\sum_{k=1}^{n} (k+1)^p = \sum_{k=2}^{n+1} k^p = S_{n+1}(p) -1 = S_n(p) +(n+1)^p -1$$
Hence finally we get the formula :
$$\color{red}{(n+1)^p -1  =\sum_{i=0}^{p-1}\binom{p}{i} S_n(i)} $$

From this it is possible to compute the sum for any $p\ge 1 $ in $ \mathbb N $.

