This isn't as hard as it looks.

Pre-requisites:

$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$

$\sum_{k=0}^{n}\binom{n}{k}=2^k.$

Note that

$\binom{n+2}{2}$

$=\frac{(n+2)!}{2!n!}$

$=\frac{(n+2)(n+1)n!}{2!n!}$

$=\frac{(n+2)(n+1)}{2}$

$=\frac{n^2+3n+2}{2}$

$=\frac{n^2+4n-n+2}{2}$

$=\frac{(4n+2)+(n^2-n)}{2}$

$=\frac{(4n+2)+[n(n-1)]}{2}$

$=\frac{4n+2}{2}+\frac{n(n-1)}{2}$

$=2n+1+\frac{n!}{2(n-2)!}$

$=n+n+1+\binom{n}{2}$

$=\binom{n}{1}+\binom{n}{n-1}+\binom{n}{n}+\binom{n}{2}$

$=\binom{n}{1}+\binom{n}{2}+\binom{n}{n-1}+\binom{n}{n}$

Adding this to

$\sum_{k=3}^{n-2}\binom{n}{k},$

we have

$\sum_{k=1}^{n}\binom{n}{k}=4095\cdots(1)$

However, it is well-known that

$\sum_{k=0}^{n}\binom{n}{k}=2^k.$

So let us add $\binom{n}{0}=1$ to both sides of $(1)$ to obtain

$\sum_{k=0}^{n}\binom{n}{k}=4096.$

$=2^{12}$

Hence $n=12\implies\frac{n^2}{2}=\boxed{72}$ , and we are done.