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.