Integrating the arithmetic series to get the sum of first $n$ squares? I only learnt today that series can be differentiated and integrated and I wondered if it is possible to apply the techniques of differential calculus to the finite arithmetic series
$$\sum_{x=0}^n x = n(n+1)/2$$
to obtain the sum of the first $n$ squares:
$$\sum_{x=0}^n x^2 = \frac{n(n+1)(2n+1)}{6}$$

Let's say we have, $$f(x)=1+x+x^2+x^3+\cdots+x^n$$
$$\frac{x^{n+1}-1}{x-1} = 1+x+x^2+x^3+\cdots+x^n$$
Differentiating,
$$\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} = 1+2x+3x^2+\cdots+nx^{n-1}$$
Multiplying both sides by $x$
$$\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2} = x+2x^2+3x^3+\cdots+nx^{n}$$
Differentiating, again
$$\frac{(-2n^2-2n+1)x^{n+1} +n^2x^{n+2} +(n+1)^2x^n-x-1 }{(x-1)^3} = 1+2^2x+3^2x^2+\cdots+n^2x^{n-1}$$
Then take the limit at $x\to1$,
and we're done.
Disclaimer: This is not at all advisable, and was done just for curiosity. I had to use WA for the differentiation, as I lack the patience
