<s>[INST] Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
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?
 [/INST] Late to the party, but I hope I am contributing something new to the answers already posted here.
We seek
$$\sum_{i=1}^Ni^2=1^2+2^2+\cdots+N^2$$
which, geometrically speaking, is equivalent to adding the area of $N$ squares with side lengths $1,2,...,N$. A visual depiction when $N=2$ is
                                                           
By adding the missing tile on the top-left corner, we may form a rectangle as follows,
                                                           
with area $2(1+2)$. Then our sum, for $N=2$, is
$$S_2=1^2+2^2=\text{Area rectangle}-\text{Area tile} = 2(1+2)-(1\cdot1).$$
To get a better idea, consider the case with $N=3$,
                             
and again, adding the missing tiles so that we form a rectangle, gives
                             
with area $3(1+2+3)$. The desired sum, for $N=3$, is
$$S_3=1^2+2^2+3^3=\text{Area rectangle}-\text{Area tiles} = 3(1+2+3)-[\underbrace{1\cdot1}_\text{blue tile}+\underbrace{1\cdot(1+2)}_\text{red tile}].$$
Drawing a few more, one may notice that the height of each tile is $1$ and the base is the sum of the bases of the previous squares. In general, we have
$$
\begin{aligned}
S_N = \sum_{i=1}^Ni^2&=N\left(\sum_{i=1}^Ni\right)-\left[1\cdot1+1\cdot(1+2)+\cdots+1\cdot(1+2+\cdots+N-1)\right]\\
& =N\left(\sum_{i=1}^Ni\right)-\sum_{i=1}^{N-1}\sum_{j=1}^{i}j=N\left(\frac{N(N+1)}{2}\right)-\sum_{i=1}^{N-1}\frac{i(i+1)}{2}\\
&=\frac{N^2(N+1)}{2}-\frac12\sum_{i=1}^{N-1}i^2-\frac12\sum_{i=1}^{N-1}i=\frac{N^2(N+1)}{2}-\frac12\sum_{i=1}^{N-1}i^2-\frac{(N-1)N}{4}\\
&=\frac{N^2(N+1)}{2}-\frac12\sum_{i=1}^{\color{red}{N}}i^2+\frac{N^2}2-\frac{(N-1)N}{4},
\end{aligned}$$
and now solving for $\sum i^2$ leads to
$$\frac32\sum_{i=1}^Ni^2=\frac{N^2(N+1)}{2}+\frac{N^2}2-\frac{(N-1)N}{4}\implies\sum_{i=1}^Ni^2=\frac{N(2N+1)(N+1)}{6}.$$
</s>