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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:

enter image description here

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

enter image description here

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!

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    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. – 2017-02-11

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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.

The proof given in a picture you show us goes exactly in this direction.

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    so basically, I am going backwards, or something like that yea? – 2017-02-11
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    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. – 2017-02-11