Can we derive the formula for summation of the series $1^{m}+2^{m}+3^{m}+...+n^{m}$ for every $m\geq2$? [duplicate]

Question

Can we derive the formula for summation of the series $1^{m}+2^{m}+3^{m}+...+n^{m}$ for every $m\geq1$ ? For e.g. $1+2+3+...+n=\frac{n(n+1)}{2}$ . Here $m=1$ . Now I also know that $1^{2}+2^{2}+3^{2}+...+n^{2}=\frac{n(n+1)(2n+1)}{6}$ . Here $m=2$ . Also again $1^{3}+2^{3}+...+n^{3}=(\frac{n(n+1)}{2})^{2}$ . Here $m=3$ . But I don't know the formulas for $1^{m}+2^{m}+...+n^{m}$ for all $m\geq4$ . How to find the formulas easily by induction? Please help me out.

Answer 1

We can use a modified form of Pascal's Triangle.

Render

$i^n=\sum_{m=1}^n a_{n,m}\binom{i+n-m}{1+n-m}$

where

$a_{n,n}=(-1)^{n-1}$

$a_{n,1}=n!$

$a_{n,m}=(n-1)(a_{n-1,m}-a_{n-1,m-1})$

Then the requesed sum will be

$\sum_{i=1}^k i^n=\sum_{m=1}^n a_{n,m}\binom{k+1+n-m}{2+n-m}$

For instance, with $n=4$ we use the recursion relation defined above to find the triangular coefficients

$a_{4,1}=24,a_{4,2}=-36,a_{4,3}=14,a_{4,4}=-1$

So

$i^4=24\binom{i+3}{4}-36\binom{i+2}{3}+14\binom{i+1}{2}-\binom{i}{1}$

and then

$\sum_{i=1}^k i^4=24\binom{k+4}{5}-36\binom{k+3}{4}+14\binom{k+2}{3}-\binom{k+1}{2}.$

When we expand the polynomials represented by the binomial terms we get

$\dfrac{24k^5+60k^4+40k^3-4k}{120}.$