Prove via induction $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$ How do you prove in induction that:
$$\frac{(n+1)^2}{2^n}\le\frac{9}{4}$$
This is what I keep getting:
Checking for $n=1$ we get $2\le\frac{9}{4}$.
Assuming it's true for $n$ and checking for $n+1$ I get this:
$$\frac{(n+2)^2}{2^{n+1}}=\frac{2(n+1)^2-n^2+2}{2\times2^n}\le\frac{9}{4}-\frac{n^2-2}{2\times2^n}\le\frac{9}{4}$$
Which is true only for $n>1$.

The simple answer is to check $n=2$ too, where you have 
$$\frac{(n+1)^2}{2^n}= \frac{(2+1)^2}{2^2} = \frac{9}{4} \le\frac{9}{4}$$ and then do the induction.
