Sum of the powers of natural numbers

[pyfgcr]

Hi everyone. I have learned that:
1+2+3+...=$\frac{n(n+1)}{2}$
1²+2²+3²=$\frac{n(n+1)(2n+1)}{6}$
I want to know what the general formula of Ʃn^a, in which n and a are natural numbers, respect to n and a.

 

[Mute]

See Faulhaber's formula (and the page about Bernoulli numbers, as they appear in the general formula).

 

[AlephZero]

Actually, a simpler generalization of
1+2+3+ ... = n(n+1)/2
is
1.2 + 2.3 + 3.4 + ... = n(n+1)(n+2)/3
1.2.3 + 2.3.4 + 3.4.5 + ... = n(n+1)(n+2)(n+3)/4
etc.

 

[pyfgcr]

Now I know. Thanks for the answer.