calculus - Proving $sum_{k=1}^n{k^2}=frac{n(n+1)(2n+1)}{6}$ without induction

I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$

It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?

Answer

$$n^{3}-(n-1)^{3}=3n^{2}+3n+1$$ $$(n-1)^{3}-(n-2)^{3}=3(n-1)^{2}+3(n-1)+1$$ $$\vdots$$ $$2^{3}-1^{3}=3(1)^{2}+3(1)+1$$

Now use telescopic cancellation.

Here are some "proof without words"(I find them more elegant):

Finally a more generalized form:$$1^{k}+2^{k}+\cdots+n^{k}=\sum\limits_{i=1}^{k}S(k,i)\binom{n+1}{i+1}i!$$ Where S(k,i) represents the Stirling number of the second kind.