# Thread: Very basic induction (sum of first n positive integers) - help needed

1. ## Very basic induction (sum of first n positive integers) - help needed

Whenever someone has a moment, can you answer this? It is the very first example in the book: http://i.imgur.com/lxbdp.jpg
(I underlined the part in red in the picture)
This is the only part I couldn't understand right from the beginning.

In the inductive step, when they are writing the formula for P(k+1), how did they get it?

On the right side all clear - they replaced k with k+1 as I would expect.

But on the left side there's still k and they added (k+1)? I don't get that.
Why didn't they replace k with k+1, like they did on the right side?

Thanks.

2. ## Re: Very basic induction (sum of first n positive integers) - help needed

Well the sum of the first k+1 positive integers is the sum of the first k positive integers plus the number (k+1).

So $P(k+1)=P(k)+(k+1)=\frac{k(k+1)}{2}+(k+1)~.$

3. ## Re: Very basic induction (sum of first n positive integers) - help needed

Strictly speaking, $P(k)$ is a proposition (equality in this case), not a number. The left-hand side of $P(k)$ is $\sum_{i=1}^k i$, or, in words, the sum of all numbers from 1 to $k$. In writing $P(k+1)$ they did replace $k$ with $k+1$ in both sides of the equality $P(k)$. The left-hand side became $\sum_{i=1}^{k+1} i$, or $1+\dots+k+(k+1)$. In words, "the sum of all numbers from 1 to $k+1$" mean the same thing.

4. ## Re: Very basic induction (sum of first n positive integers) - help needed

Thanks, that does make it clear. Also explains several other problems I was having trouble with.
I was thinking in terms of "replace k by k+1 in the expression", but now I see that I need to treat P(k+1) as a proposition, i.e. go back to the part where the proposition was defined (in text), and put k+1 there.
Thanks again, I very much appreciate your help. That was my first post, and I'm feeling like this place is going to be a great asset.