Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.

Question

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 $n \in \mathbb{N}$.

Answer 1

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).

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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. $\Box$

Answer 2

Consider any natural number $r$. You have $$r^3-(r-1)^3=3r^2-3r+1.$$

Now telescope it: $$ 1^3-0^3=3-3+1 $$ $$2^3-1^3=3\cdot2^2-3\cdot2+1 $$ $$\vdots $$ $$ n^3-(n-1)^3=3n^2-3n+1 $$ Now add, and see them cancel out. You are left with $$n^3=3(1^2+2^2+\cdots+ n^2)-3(1+2+3+\cdots+n)+n$$ You must know $$ 1+2+3+\cdots+n=\frac{n(n+1)}{2}. $$ Plug it in, and you get the answer. Also, please see that this method works even for $\sum r^4,r^5,\cdots$. I have tried it out. All you need is the sum of its previous powers.

Answer 3

Your inductive assumption is such that the formula marked $\color{red}{\mathrm{red}}$ (several lines below) holds for $i=k$:

$$\sum^{i=k}_{i=1} i^2=\frac{k(k+1)(2k+1)}{6} $$

You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^2=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}$$

To do this you cannot use: $$\sum^{i=k}_{i=n} i^2=\color{red}{\frac{n(n+1)(2n+1)}{6}} $$ as this is what you are trying to prove.

So what you do instead is notice that: $$\sum^{i=k+1}_{i=1} i^2= \underbrace{\frac{k(k+1)(2k+1)}{6}}_{\text{sum of k terms}} + \underbrace{(k+1)^2}_{\text{(k+1)th term}}$$ $$\sum^{i=k+1}_{i=1} i^2= \frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$$ $$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)\left(k(2k+1)+6(k+1)\right)}{6}$$ $$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)(2k^2+\color{green}{7k}+6)}{6}=\frac{(k+1)(2k^2+\color{green}{4k+3k}+6)}{6}=\frac{(k+1)\left(2k(k+2)+3(k+2)\right)}{6}=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}\quad \forall \space k \in \mathbb{N}$$

Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.