# Prove that $2*(1+2+…n) = n*(n+1)$

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.

• Don't really know how formal you are trying to get, but $a + 1 = S(a)$, can you prove this? Then, how can split up $2 \times (1 + 2 + \ldots + n + S(n))$? Using induction, what can you assume about what you split up? – Dair Jan 17 '17 at 1:03
• @Dair In a different problem, I did prove $a+1=S(a)$. Are you referring to $2 \times (1 + 2+... + n + n+1)$? I also said that $2 = S(1)$, but I'm not sure how to tie those two together to give me the RHS of the equation. – Kosta Jan 17 '17 at 1:07
• $2 \times (1 + 2 + \ldots + n + S(n)) = 2 \times (1 + 2 + \ldots + n) + 2 \times S(n)$, what can you say about $2 \times (1 + 2 + \ldots + n)$? – Dair Jan 17 '17 at 1:09

## 1 Answer

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.

• Oh, duh. So continuing on with the RHS, $(n+S(1)) \times S(n) = (n+1+1) \times S(n) = (S(n)+1) \times S(n)$ So we've shown that the equation also holds true for its successor, therefore the equation is true for all $n$ in the natural numbers. Is this correct? – Kosta Jan 17 '17 at 1:34
• Pretty much. The main things are that you prove $P(0)$, and you prove $P(n) \rightarrow P(S(n))$, and then the Axiom of Induction says that $P(n)$ is true for all $n$ - and you have indeed done so. – ConMan Jan 17 '17 at 4:00