On Mathematical Induction

OK, you proved that $$1 + 2 + \dots + n = \frac{n(n+1)}{2}$$ without explicitly using induction.

But your proof relies, often implicitly, on a lot other results in arithmetic. For instance, that addition is commutative. And how do you prove commutativity? Trust me, you need induction (see here for instance). Maybe you can find a proof of commutativity that do not use induction explicitly, but it necessarily uses other lemmas that rely on induction.

So, also your proof relies (indirectly) on induction.

The "necessity" of induction in arithmetic to prove non-trivial properties of natural numbers has been formalized for the first time by Peano. If you are interested to know what you can prove in arithmetic without never using induction (not even implicitly), see Robinson arithmetic.

I remember when I substitute taught high school and had to try to explain what a "metaphor" was. The students couldn't help but think it had to be something confusing and complicated and didn't get what the rules were. In actuality the difficulty in knowing what a "metaphor" was is because it is such a basic concept we use them all the time.

Induction is similar.

We know that $1+ 2+3+ 4+5 = 5+4+3+2+1$ but how do we know that $1+2+..... + n = n+(n-2) + .... +1$ for all possible $n$?

Well, addition is commutative weren't we told that? Well, no we were told that it is commutative for two elements that $a+b = b+a$ but we were never told $a+b+c = c+b+a$. In fact we know addition is not commutative infinitely as $1+(-1) + 1+(-1)+..... $ can not be rearranged to get put five of the positive ones in front and intersperse the rest throughout to get an answer of $5$.

But can't we infer that if it's true for two in must be true for any $n$ by doing it just two at a time until we get to $n$? Well, yes we can, but what do you think we call that concept of "doing it multiple times until we get to $n$". The word for that is.... induction.

Any hoo.... so the proof

$S = 1 + 2 + ....+n$ so $2S = (1+n) + (2+n-1) + .... + (n+1)=n(n+1)$

is valid. But it is also a prove by induction. Not heavy-metal steel scaffolding induction but an implicit "induction is in the air you breath" induction. (Hey! That was a metaphor!)

...

So I guess my point is, although we don't need formal and rigorous and "scary" induction, we shouldn't consider "induction" to be the heavy structure it seems.

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To answer your question "Is the use of mathematical induction inevitable in this situation? " I'm not sure. But I think if we examine what we think "induction" is, it's not such an all or nothing question. I'm inclined to say yes, induction is inevetible but ... I'm not sure.

Start with

$S=1+\dots+n$

$S=n+\dots+1$

I'd suggest that, then, "adding both equations," we write $$2S = \underbrace{(n+1) +\cdots + (n+1)}_{n \text{ times}}$$ Just being more explicit, so we do, in fact then get $2S = n(n+1)$, and dividing by $2$ gives us $$S= \frac{n(n+1)}{2}.$$

But note that this appeals to intuition, and all the "$+\cdots +$" amount to a bit of hand-waving. This is @fleablood's point in the comments. It is a great start for establishing a proposition worth testing by inductive proof.