I'm attempting to prove the following.
Prove that $2\times (1+2+...+n) = n \times (n+1) $ for all natural numbers $n$.
We're only allowed the use of Peano's axioms and the induction method to solve this proof. I've shown the basis step of the induction easy enough.
Assume $n=1$ and $S(n)$ defines the successor of $n$. Then, $$ 2 \times (1) = 1 \times (1+1) \\ 2 = 1 \times S(1) \\ 2 = S(1) $$
My problem is proving this is true for all $n$. We know that by Peano's second axiom, $n$ must have a successor $S(n)$. We can then say that
$$ 2 \times (1+2+...+n+S(n)) = S(n) \times (S(n)+1) $$
From here, I began operating on the right-hand side in an attempt to find the left-hand side. But, I keep finding myself going in circles with the algebra. I've tried this problem 4 different ways and cannot seem to find where to use the induction hypothesis that allows me to finish the proof.
Any hints would be greatly appreciated, thank you.
You want to prove that, given the proposition $P(n) \equiv 2 \times (1 + 2 + \ldots + n) = n \times (n + 1)$, that you have the inductive step $P(n) \rightarrow P(S(n))$. So, what you want to do, is start from $P(n)$ and manipulate it to look more like $P(S(n))$. So we have:
$$\begin{eqnarray}& 2 \times (1 + 2 + \ldots + n) & = & n \times (n + 1)\\ & 2 \times (1 + 2 + \ldots + n) + 2 \times S(n) & = & n \times (n + 1) + 2 \times S(n) & \mbox{adding S(n)}\\ & 2 \times (1 + 2 + \ldots + n + S(n)) & = & n \times (n + 1) + 2 \times S(n) & \mbox{by associativity}\\ & & = & n \times S(n) + 2 \times S(n) & \mbox{(1)} \\ & & = & (n + 2) \times S(n) & \mbox{distributivity} \end{eqnarray}$$
and from there you should be able to rearrange things as necessary. Note that I'm assuming you already have an appropriate theorem that identifies $n + 1 = S(n)$, which you will have to use again when you complete the proof. Without that, it might be a little trickier.