An airplane crashes on an island in the middle of nowhere. The $n$, ($n \geq 1$) passengers all end up apart from everyone else. We call each passenger a group. The passengers start wandering around. As each group meets another group everyone from one group shakes hands with everyone from the other group, and then they form a new larger group. Eventually everyone meets and they form one large group of size $n$.
Assume the formula for the the handshakes is $an^2 + bn + c$.
So I know the base case is $n = 2$ cause one person can't have a handshake That would be $a^4 + 2b + c$.
For the inductive hypothesis, do I assume for $n \geq 2$?
To continue withe the direction Arthur mentioned, see that the total number of handshakes will be $$f(n) = \frac{n(n-1)}{2}$$
This is by a combinatorial argument rather than an induction proof. See that everyone ultimately will shake hands with everyone else. So there are $n$ choices for the first person involved with the handshake, and $n-1$ people for the other. Doing this, we over count by a factor of $2$ so we must divide by this factor to get the desired result.
But if you would also like an induction argument for why this formula works, continue reading:
Plugging in $n=1$ into $f(n)$ does result in $0$ handshakes, as it should.
You can then assume it's true for $k\ge 1$ and the goal is to show that it is true for $k+1$.
The number of handshakes of $k+1$ people can be looked at as the number of handshakes that involve the $(k+1)$th person and the number of them that don't.
There will be $k$ other people this person needs to handshake with, and there are the other $f(k)$ handshakes that did not involve them so we have $$k + f(k) = k + \frac{k(k-1)}{2} = \frac{2k + k^2 - k}{2} = \frac{(k+1)(k)}{2} = f(k+1) $$
as desired.