Basic Mathematical Induction

I'm not quite sure how to approach this question.

I need to prove that for

$$n\ge1$$

$$1^2+2^2+3^3+\dots+n^2=\frac16n(n+1)(2n+1)$$

Do I just plug $1$ and see if

$$\frac16(1)((1)+1)(2(1)+1) = 1^2\text{ ?}$$

Answer

Well what we want to do with induction is to show that for the set $S$ that satisfies our desired property, $S = \mathbb{N}$. We can do this on the naturals by a very special property. Every as part of the Peano axioms, we know that $\forall x \in \mathbb{N}$, $x$ has a successor also in $\mathbb{N}$ in the form $\text{s} (x) = x+1$. So if we can show that if the base natural $1$ holds and that our property holds for an arbitrary $x$ then it holds for its successor, we prove the property over the naturals. This works because if $1,x,\text{s}(x) \in S$ we can let $1$ be our $x$ and then we get that it holds for s$(1)$ so it holds for s$($s$(1))$ and so on.

So to the problem at hand. Let $S = \left\{ x \in \mathbb{N} : 1^2 + 2^2 + ... + x^2 = \frac{x(x+1)(2x+1)}{6} \right\}$. First we show the base case $1 \in S$. So,

$$1^2 = 1 = \frac{1(1+1)(2*1+1)}{6} = \frac{6}{6} = 1$$ therefore $1 \in S$.

Next we form our inductive hypothesis and assume that some natural $k \in S$ which implies $1^2 + 2^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}$. We will use this to make our inductive step and show $k+1 \in S$. So,

$$\begin{align*} 1^2 + 2^2 + ...+k^2 + (k+1)^2 =& \frac{k(k+1)(2k+1)}{6} + (k+1)^2\\ =&\frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}\\ =&\frac{(k+1)[k(2k+1) + 6(k+1)]}{6}\\ =&\frac{(k+1)[2k^2+k + 6k+6)]}{6}\\ =&\frac{(k+1)[2k^2+7k+6)]}{6}\\ =&\frac{(k+1)[(k+2)(2k+3)]}{6}\\ =&\frac{(k+1)[(k+1+1)(2k+2+1)]}{6}\\ =&\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}\\ \end{align*}$$ Which is what we trying to show! $\blacksquare$

Side note: I should probably conclude that we showed that when $k \in S$ that $k+1 \in S$ which means that $S = \mathbb{N}$.