Formulating Sum as Function of n

Question

Can someone please help me rewrite the following as a function of $n$ in closed form?

$$\sum_{k=0}^{n^2-2} {k+2 \choose 2} {n^2-2 \choose k}$$

Thank you!

Answer 1

Let $n^2-2=N$, Then $$S=\frac{1}{2} \sum_{k=0}^{N} (k^2+3k+2) {N \choose k}~~~~(1)$$ Use the standard results $$\sum_{k=0}^{N} {N \choose k}=2^N, \sum_{k=0}^{N} k {N \choose k}=N 2^{N-1}, \sum_{k=0}^{N} k^2 {N \choose k}= N(N+1) 2^{N-2}.$$ in (1), you get the required result: $$S=(n^4+3n^2-2)~ 2^{n^2-5}$$

Answer 2

Using $$x^2(1+x)^{n^2-2}=\sum_{r=0}^{n^2-2} {{n^2-2} \choose r} x^{r+2}$$ Differentiate twice we get $$(1+x)^{(n^2-2)}(x(n^2-3)(2+n^2x)+2(1+x)(1+n^2x))=\sum_{r=0}^{n^2-2}(r+2)(r+1){{n^2-2} \choose r}$$ Now put x=1 on both sides and divided by 2 on both sides $$\sum_{r=0}^{n^2-2} \frac{(r+2)(r+1)}{2} {{n^2-2} \choose r}=2^{n^2-5}(n^4+3n^2-2)$$