# Simple Question about proof by Mathematical Induction

This might be a silly question but I have never had proofs at school and I get stuck at the uni especially in a foreign country. So here goes...

The following needs to be proven by Mathematical Induction:

This is an example in my Script so here is the solution as well:

Step 1: (I will skip this, it's just n=1 and it is correct)

Step 2: $A(n) \implies A(n+1)$

Well, I understand what is written there, but, when I 'follow the instructions' of the Mathematical Proof I get something like this (I basically put $n+1$ instead of $n$):

$$\sum_{k=1}^{n+1} k = \frac {(n+1)*(n+2)}{2} = (n+1) + \sum_{k=1}^{n} k = \frac {(n+1)*(n+2)}{2}$$

$$\sum_{k=1}^{n} k = \frac {(n+1)*(n+2)}{2} - (n+1) =\frac {n*(n+1)}{2}$$

Is what am I doing also proof and equivalent to the other solution given above, or am I completely lost?

Thanks!

• You do need to go the right way to use induction. However, using reversible steps to work backwards can, in some cases, show how the induction can be carried through. Feb 11 '17 at 19:56

Assuming $$\sum_{k=1}^nk=\frac{n(n+1)}{2}$$ you should check that $$\sum_{k=1}^{n+1}k=\frac{(n+1)(n+2)}{2},$$ but you start from this equation. You must not assume the assertion of the theorem.
• What you are calling "going backwords" is the induction hypothesis. In the case in question this is the equation $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$ Of course, there are also more advanced techniques of induction proofs, but this which we discuss is the most typical one. Feb 11 '17 at 19:51