summation - Why does he need to first prove the sum of the $n$ integers and then proceed to prove the concept of arithmetical progression?

I'm reading What is Mathematics, on page 12 (arithmetic progression), he gives one example of mathematical induction while trying to prove the concept of arithmetic progression. There's something weird here: he starts by proving that the sum of the first $n$ integers is equal to $\frac{n(n+1)}{2}$ by giving this equation:

$$1+2+3+\cdots+n=\frac{n(n+1)}{2}.$$

Then he adds $(r+1)$ to both sides:

$$1+2+3+\cdots+n+(r+1)=\frac{n(n+1)}{2}+(r+1).$$

Then he solves it:

$$\frac{(r+1)(r+2)}{2}$$

Now it seems he's going to prove the arithmetical progression: He says that this can be ordinarily shown by writing the sum $1+2+3+\cdots+n$ in two forms:

$$S_n=1+2+\cdots+(n-1)+n$$

And:

$$S_n=n+(n-1)+\cdots+2+1$$

And he states that on adding, we see that each pair of numbers in the same column yields the sum $n+1$ and, since there are $n$ columsn in all, it follows that:

$$2S_n=n(n+1).$$

I can't understand why he needs to prove the sum of the first $n$ integers first. Can you help me?

Thanks in advance.

EDIT: I've found a copy of the book on scribd, you can check it here. This link will get you in the page I'm in.

EDIT:
I kinda understand the proofs presented in the book now, but I can't see how they are connected to produce a proof for arithmetic progression, I've read the wikipedia article about arithmetic progression and this $a_n = a_m + (n - m)d$ (or at least something similar) would be more plausible as a proof to arithmetic progression - what you think?

Answer

He is giving two different proofs, one by a formal induction, and the other a more intuitive one. Good idea, two proofs makes the result twice as true! More seriously, he is probably taking this opportunity to illustrate proof by induction.

It is important to know the structure of a proof by induction. In order to show that a result holds for all positive integers $n$ one shows (i) that the result holds when $n=1$ and (ii) that for any $r$, if the result holds when $n=r$, then it holds when $n=r+1$.

(i) is called the base step and (ii) is called the induction step.

Almost always, the induction step is harder than the base step.

Here is how the logic works. By (i), the result holds when $r=1$. By (ii), because the result holds for $n=1$, it holds when $n=2$ (we have taken $r=1$). But because the result holds for $n=2$, it holds when $n=3$ (here we have taken $r=2$). But because the result holds when $n=3$, we can argue in the same way that the result holds when $n=4$. And so on.

In our example, suppose that we know that for a specific $r$, like $r=47$, we have

$$1+2+\cdots+r=\frac{r(r+1)}{2.}$$
We want to show that this forces the result to hold for the "next" number.
Add $(r+1)$ to both sides. We get
$$1+2+\cdots +r+(r+1)=\frac{r(r+1)}{2}+(r+1).$$
Now we do some algebraic manipulation:
$$\frac{r(r+1)}{2}+(r+1)=\frac{r(r+1)+2(r+1)}{2}=\frac{(r+1)(r+2)}{2},$$
which is what the formula we are trying to prove predicts when $n=r+1$. We have taken care of the induction step. The base step is easy. So we have proved that $1+2+\cdots+n=\frac{n(n+1)}{2}$ for every positive integer $n$.

Remark: Here is another way to think about the induction, one that I prefer. Suppose that there are positive integers $n$ for which the result is not correct. Call such integers $n$ bad. If there are bad $n$, there is a smallest bad $n$. It is easy to see that $1$ is not bad.

Let $r+1$ be the smallest bad $n$.

Then $r$ is good, meaning that $1+2+\cdots+r=\frac{r(r+1)}{2}$. Now we argue as in the main post that $1+2+\cdots +r+(r+1)=\frac{(r+1)(r+2)}{2}$. That shows that $r+1$ is good, contradicting the assumption that it is bad.