It's well known that if you want to prove a theorem that involves $n$ elements (terms,sets,variables,sequences,etc.) you should think about induction.
But is there another way to proof a theorem with that characteristics?
For instance:
What if I prove that the result holds for the base case $(n=1)$ and then for $n=2$ and $n=3.$ Could I conclude that the result holds for the $n$ case$?$
On
Nope. You have only proved it for $n=1,2,3$.
In general, you can only prove that kind of theorem by induction.
If you seem to avoid induction, you will have to use theorems that themselves are proved by induction.
For example, consider this proof that, if $s(n) =\sum_{k=1}^n k $ then $s(n) =\frac12 n(n+1) $.
Changing the order of summation, $s(n) =\sum_{k=1}^n (n+1-k) $. Therefore
$\begin{array}\\ 2s(n) &=\sum_{k=1}^n k+\sum_{k=1}^n (n+1-k)\\ &=\sum_{k=1}^n (k+n+1-k)\\ &=\sum_{k=1}^n (n+1)\\ &=n(n+1) \end{array} $
Look - no induction!
However, the definition of $\sum_{k=1}^n$ and the manipulation of the summation indices both depend on induction to prove that they can be used as stated.
Another way to see this is to note that the Peano axioms which define the integers only have one way to do the definition: induction.
The thing is you can't guarantee it'll hold for the $n$ case. One example for instance are the Borwein integrals, who follow a pattern up to $n=14$ then fail at the next iteration. This is the worry with many conjectures out there like the Collatz conjecture that's been checked for $ 5×2^{60}$ iterations, but could technically fail at one point, until we prove it always holds (induction doesn't necessarily work for Collatz, but you get my drift.)
Induction doesn't "right off the bat" prove it works for all integers, you just show it works for a base case and induction guarantees it'll hold for the next integer -- and as a result, the one after, and one after, etc.