On Mathematical Induction Let's suppose we are asked to prove $1+2+\ldots+n=\frac{n(n+1)}{2}$, for a natural number $n$. Is the use of mathematical induction inevitable in this situation? For instance, what makes the following proof not mathematically sound?
$S=1+\dots+n$
$S=n+\dots+1$
Adding both sides of the two equations gives $2S=\overbrace{(n+1)+\dots+(n+1)}^{n \text{ times}}$, and dividing through by $2$ yields the result.
 A: 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.
A: 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.
...
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.
A: 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.
A: As has been pointed out in some of the answers, the ellipsis is an implicit induction. It means, and so on, the one following from the other. And that is pretty much what induction means. You show that P(0) is true, and that P(n) implies P(n+1). It is a licence to make a quick and simple argument to get to any n, and so we assert that it means that the result is true for all n. 
To expand on this, consider the assertion in another another answer that the sum of 1st powers $\sum_{i=0}^n i$ must be a polynomial of the 2nd order. And the (apparent) implicit principle that $\sum_{i=0}^n i^m$ must be a polynomial in $n$ of order $m+1$. From where comes this idea?
I think that it is reasonable to say that the idea comes from the approach attributed to Gauss, that (1+2+...+n) + (n + ... + 2 + 1) = (n+1) + (n+1) + ... + (n+1). This leads to the answer n(n+1)/2. Which is a polynomial of 2nd order, for a 1st order sum.
Note: the ellipsis there is an assertion that there is a regular pattern that will continue for the length of the sum. This is the inductive assumption. The fact that this can typically be seen intuitively by most people does not stop it from being induction.
There is also an earlier version of proof by induction that is a bit more careful, but swaps proof by exhaustion for proof by contraction. That is, if P(n) is false then P(n-1) is also false (that bit has to be proved), and so eventually P(0) must be false, but it is not (that bit has to be proved) and so p(n) must actually be true. This version involves no and-so-on to infinity. (And beyond)?

And now for the 2nd and stranger part. 
The assertion is that $\sum_{i=0}^n i^m$ is a polynomial in $n$ of order $m+1$. We can test this by trying $\sum_{i=0}^n i = ai^2+bi+c$. If this is true, then it must be true in particular for $n=0,1,2$. This gives 3 linear equations in the 3 unknowns a,b,c and we work out their values and get $n(n+1)/2 $. 
But, to prove that this formula works in general we are pretty much back to showing that P(n) implies P(n+1). This can  be done, and the process can be used to prove the general assertion for m=1,2,3, or any small finite. 
The heuristic - to find a viable formula for $\sum_{i=0}^n i^m$ that can be proved by induction, use a general polynomial of order $m+1$ - is very good advise, and to use it you do not need to have a proof of it.
But, to show it works now we need to prove that this is true of all m.
One approach would be to prove that if the $\sum_{i=0}^n i^m$ is an $m+1$ order polynomial then $\sum_{i=0}^n i^{m+1}$ is an $m+2$ order polynomial. 
The formula is known as Faulhaber's formula, although Bernoulli was also involved. Some proofs assert to prove the result from complex number manipulation, however they use the exponential generating function, which is a sum, and to prove the properties of these sums, you need, ultimately induction. 

So, is this really induction? If you use in a prove a theorem that might have to be proved by induction, but avoids explicit use of induction, then is this or is this not a proof that uses induction?
This comes to the part of the question that is subjective. 
Do you need to explicitly use induction to prove the sum of squares, for example. No. You can use Faulhaber's formula, and be done with it. But, then Faulhaber's formula is an assumption that is much much more general that the thing being proved. I could prove that
1+2+...+n=n(n+1)/2 from the assumption that the sum of an arithmetic series is the average of the first and last times the number of items. This is not induction, but it is also using something more general to prove a specific. 
The point is that there is no method to prove the result from first principles without using induction. And so, the answer to the question is subjective.
Is this all? No.
Because the method of induction can be proved from the well-ordering principle. Meaning that any non empty set of positive integers has a lowest element. And so, I would not strictly be using induction. However, the basic work required to do this is identical to the proof by induction - it is just a way of rephrasing it.
A: Let $S_n$ be the sum up to $n$. We have
$$S_n-S_{n-1}=n, $$ which is a polynomial of the first degree. Hence $S_n$ must be a polynomial of the second degree, let $an^2+nb+c$.
Now
$$a(n-(n-1)^2)+b(n-(n-1))+c(1-1)=a(2n-1)+b=n\iff a=b=\frac 12.$$
Finally, as $S_0=0$, we have $c=0$.
This takes no induction.
