algebra precalculus - An incorrect method to sum the first $n$ squares which nevertheless works

Start with the identity

$\sum_{i=1}^n i^3 = \left( \sum_{i = 1}^n i \right)^2 = \left(\frac{n(n+1)}{2}\right)^2$.

Differentiate the left-most term with respect to $i$ to get

$\frac{d}{di} \sum_{i=1}^n i^3 = 3 \sum_{i = 1}^n i^2$.

Differentiate the right-most term with respect to $n$ to get

$\frac{d}{dn} \left(\frac{n(n+1)}{2}\right)^2 = \frac{1}{2}n(n+1)(2n+1)$.

Equate the derivatives, obtaining

$\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)$,

which is known to be correct.

Is there any neat reason why this method happens to get lucky and work for this case?

Answer

Let $f_k(n)=\sum_{i=1}^n i^k$. We all know that $f_k$ is actually a polynomial of degree $k+1$. Also $f_k$ can be characterised by the two conditions:

$$f_k(x)-f_k(x-1)=x^k$$ and $$f_k(0)=0.$$ Differentiating the first condition gives $$f_k'(x)-f_k'(x-1)=k x^{k-1}.$$ Therefore the polynomial $(1/k)f_k'$ satisfies the first of the two conditions that $f_{k-1}$ does. But it may not satisfy the second. But then $(1/k)(f_k'(x)-f_k'(0))$ does. So $$f_{k-1}(x)=\frac{f_k'(x)-f_k'(0)}k.$$

The mysterious numbers $f_k'(0)$ are related to the Bernoulli numbers, and when $k\ge3$ is odd they obligingly vanish...