Question in derivation of formulae for sum to $n $ terms of a series I have been learning that the sum of the squares of the first $n $ natural numbers is given by $$\Sigma= \frac {1}{6} n (n+1)(2n+1) $$ while the sum of the cubes of the first  $n $ natural numbers is given by: $$\left[ \dfrac {n (n+1)}{2} \right] ^2$$ I know that these can be verified by induction process, but can anyone explain how these were derived for the first time, and how similar formulae are derived? In derivation, one cannot use mathematical induction, so which process is used to derive such formulae?
Simply said, can anyone show how these formulae are derived, and not verified?
 A: You can use: $\sum_{k=0}^{n} k^{3} = \sum_{k=0}^{n} (k+1)^{3}- (n+1)^{3}$, expand the sum and see that o term cubic cancel and you have a sum in function of $\sum k$. You can use this metod for sum of $k^{4}$ ... etc.
A: $\sum_\limits{i=1}^n i^k$
One way to do this is to assume that there exists a polynomial in terms of $n$ that equals the series.
And then $p(n+1) - p(n) = (n+1)^k$ and $p(1)= 1$ allows you to solve for the coefficients.
This is basically the same algebra of the induction proof.
An alternative is to construct a telescoping series.  e.g.
$\sum_\limits{i=1}^n i^{3}-(i-1)^{3} = n^3$
Multiplying it out gives.
$\sum_\limits{i=1}^n (3i^2 - 3i +1) = n^3$
$\sum_\limits{i=1}^n i^2 = $$\frac 13 n^3 +  \sum_\limits{i=1}^n i - \frac 13 \sum_\limits{i=1}^n 1\\
\frac 13 n^3 +  \frac 12(n)(n+1) - \frac 13 n\\$
And then there are some elegant "proof without words" proofs.  Which again give nice poofs that these series sum to what they claim to sum to, but are probably not the best starting points.
