Sum of powers.

What is \sum_{j=1}^n j^k? – this is a particular questions in mathematics. In this post, I would like to work this question out.

First I define some function P_k(z), defined as

\displaystyle P_k(z) = \prod_{j=1}^k (z+j)

Note that P_k(0)=k! and P_k(-1)=0.

It is clear that

\displaystyle P_{k+1}(z) - P_{k+1}(z-1) = \prod_{j=1}^{k+1} (z+j) - \prod_{j=0}^k (z+j) = (k+1) P_k(z)

It is also clear that

\displaystyle \sum_{z=0}^n \left\{ P_{k+1}(z) - P_{k+1}(z-1) \right\} = P_{k+1}(n)

Then we obtain that

\displaystyle \sum_{z=0}^n P_k(z) = \frac{1}{k+1} P_{k+1}(n).

For example

\displaystyle \sum_{z=0}^n (z+1)(z+2) = \frac{1}{3} (n+1)(n+2)(n+3).

Then

\displaystyle \sum_{z=0}^n \left\{ z^2 + 3(z+1) - 1 \right\} = \frac{1}{3} (n+1)(n+2)(n+3).

So

\displaystyle \sum_{z=0}^n z^2  = \frac{1}{3} (n+1)(n+2)(n+3) - \frac{3}{2} (n+1)(n+2) + (n+1).

Therefore

\displaystyle \sum_{z=0}^n z^2  = \frac{1}{6} n (n+1) (2n+1)

Hope you enjoyed this post!

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