<s>[INST] The pattern in mathematical induction proofs When given a statement to be proven by mathmatical induction
the statement tends to look like this
$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
so going about the proof.
1) Prove the base case
$\frac{1\times(1+1)}{2} = 1$
2) Prove the inductive case
Assume 
$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
Now add $n+1$ to both sides of the equation.
\begin{align}
1 + 2 + 3 + \dots + n + (n + 1) & = \frac{n(n+1)}{2} + (n + 1) \\
& =  \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \\
& =  \frac{(n + 1)(n + 2)}{2} \\
& =  \frac{(n + 1)((n + 1) + 1)}{2} 
\end{align}
Thus the proof is complete.  Now my question is, can we just remove the $1 + 2 + 3 + \dots + n$ side of the equation, and put anything there? for example, if we used $f(n)$, Then the proof would look like this.
1) Prove the base case
$f(1) = \frac{1\times(1+1)}{2} = 1$
2) Prove the inductive case
Assume 
$f(n) = \frac{n(n+1)}{2}$
Now add $n+1$ to both sides of the equation.
\begin{align}
f(n) + (n + 1) & = \frac{n(n+1)}{2} + (n + 1) \\
& =  \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \\
& =  \frac{(n + 1)(n + 2)}{2} \\
& =  \frac{(n + 1)((n + 1) + 1)}{2} 
\end{align}
Thus we now know $f(n) = \frac{n(n+1)}{2}$.  But what pattern represents $f(n)$?
it's not immediately obvious that $f(n)$ is the sum of the first $n$ positive integers.
 [/INST] You are making the assumption that $f(n) = \text{ sum of the first } $n$ \text{ integers}$ during your induction. When you evaluate $f(1)$ you make this assumption in how you evaluate. And in your induction step you mean to assert:
$$f(n+1) = f(n) + (n+1)$$
before going on with your proof. Implicit in this assumption is a recursive definition of $f$ that forces $f$ to be the sum of the first $n$ integers. That is, there is exactly one function on the natural numbers such that $f(1) = 1$ and $f(n+1) = f(n) + (n+1)$, and that function is the one that sums the first $n$ integers. Indeed there are many ways to define this function, i.e. $f(n) = n(n+1)/2$, but you can prove that they are all the same function.
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