# Sum of First nn Squares Equals n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}

I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:

I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?

## Answer

Another way (by Euler, I think), from the geometric sum:

$$1+x+x2+⋯+xn=xn+1−1x−11 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides and multiply by $$xx$$:

$$x+2x2+3x3+⋯+nxn=nxn+2−(n+1)xn+1+x(x−1)2x + 2 x^2 + 3 x^3 + \cdots + n x^{n} = \frac{n x^{n+2}-(n+1) x^{n+1} +x}{(x-1)^2} \tag{2}$$

Differentiating once more, we get on the LHS

$$1+22x+32x2+⋯+n2xn−11 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $$x=1x=1$$ gives our desired sum $$∑nk=1k2\sum_{k=1}^n k^2$$. Hence, we just need to compute the derivative on the RHS in $$(2)(2)$$ (eg, with L’Hopital rule) and evaluate it at $$x→1x \to 1$$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here’s a concrete computation, applying the binomial theorem on the RHS of $$(1)(1)$$ and doing a series expansion around $$x=1x=1$$. Let

\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}

Deriving, multiplying by $$xx$$ and deriving again: $$(g'(x) \, x)’={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$
which evaluated at $$x=1x=1$$ gives the desired answer:
$${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} {n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6}$$

We can apply the same procedure for higher powers. For example:
$$\sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6 \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

Attribution
Source : Link , Question Author : Nathan Osman , Answer Author : Community