<s>[INST] Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
My work: So I think I have to do a proof by induction and I just wanted some help editing my proof.
My attempt:
Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. 
Then $$P(1)=1^2=\frac{1(1+1)(2+1)}{6}$$
$$1=\frac{6}{6}.$$
So $P(1)$ is true.
Next suppose that $P(k)=1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$ for $k \in \mathbb{N}$. Then adding $(k+1)^2$ to both sides of $P(k)$ we obtain the following:
$$1^2+2^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
$$=\frac{2k^3+3k^2+k+6(k^2+2k+1)}{6}$$
$$=\frac{2k^3+9k^2+13k+6}{6}$$
$$=\frac{(k^2+3k+2)(2k+3)}{6}$$
$$=\frac{(k+1)(k+2)(2k+3)}{6}$$
$$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
$$=P(k+1).$$
Thus $P(k)$ is true for $k \in \mathbb{N}$.
Hence by mathematical induction, $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true for $n \in \mathbb{N}$.
 [/INST] I am going to provide what I think is a nice way of writing up a proof, both in terms of accuracy and in terms of communication. You be the judge(s).

Claim: For $n\geq 1$, let $S(n)$ be the statement
$$
S(n) : 1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.
$$
Base step $(n=1)$: The statement $S(1)$ says $1^2=1(2)(3)/6$ which is true.
Inductive step $(S(k)\to S(k+1))$: Fix some $k\geq 1$ and suppose that
$$
S(k) : 1^2+2^2+3^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}
$$
holds. To be shown is that
$$
S(k+1) : 1^2+2^2+3^2+\cdots+k^2+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}
$$
follows. Starting with the left-hand side of $S(k+1)$,
\begin{align}
\text{LHS} &= 1^2+2^2+3^2+\cdots+k^2+(k+1)^2\tag{definition}\\[1em]
  &= \frac{k(k+1)(2k+1)}{6}+(k+1)^2\tag{by $S(k)$}\\[1em]
  &= (k+1)\left[\frac{k(2k+1)}{6}+(k+1)\right]\\[1em]
  &= (k+1)\frac{k(2k+1)+6(k+1)}{6}\\[1em]
  &= (k+1)\frac{2k^2+k+6k+6}{6}\\[1em]
  &= (k+1)\frac{2k^2+7k+6}{6}\\[1em]
  &= (k+1)\frac{(k+2)(2k+3)}{6}\\[1em]
  &= \frac{(k+1)(k+2)(2(k+1)+1)}{6}\\[1em]
  &= \text{RHS},
\end{align}
the right-hand side of $S(k+1)$ follows. This completes the inductive step. 
Thus, by mathematical induction, for every $n\geq 1, S(n)$ is true. $\Box$
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