Intuition to faulhabers sum of k-th power of n first integrals

Solution 1:

As far as I know it is just an assumption; that is, this is a proof that if $\sum_{k=0}^nk^2$ is a polynomial in terms of $n$, then it is $n(n+1)(2n+1)/6$. That this formula is actually correct can then be proved through induction.

I prefer a slightly different approach to this, which I think is a little more intuitive if you’re familiar with binomial coefficients. We will use the identity $\sum_{m=k}^n\binom{m}{k}=\binom{n+1}{k+1}$ (this is easy to prove with induction and the formula for binomial coefficients).

\begin{align} 2\binom{n}{2} &= n^2-n \\ 2\sum_{k=2}^n\binom{k}{2} &= 2\binom{n+1}{3} \\ &= \sum_{k=2}^nk^2-\sum_{k=2}^nk \\ &= \sum_{k=1}^nk^2-\sum_{k=1}^nk && 1^2=1 \\ &= \sum_{k=1}^nk^2-\binom{n+1}{2} && k=\binom{k}{1} \\ \sum_{k=1}^nk^2 &= 2\binom{n+1}{3}+ \binom{n+1}{2} \end{align}

The same method can be used for any power, and using the formula for binomial coefficients shows that this is identical to the polynomial form.