summation binomial theorem: express this sum ∑k=2n(nk)3n−k2(−1)2k+3∑k=2n(kn)32n−k(−1)2k+3 in the binomial form (x+y)n(x+y)n

[Qwertyuiop[]]

Hi, I want to express this sum [imath]\sum_{k=2}^{n} \binom{n}{k} 3^\frac{n-k}{2} (-1)^{2k+3}[/imath] in the binomial form [imath](x+y)^n[/imath]. So i simplified the expression and got here but not sure if it's correct, [imath]- \sum_{k=2}^{n}\binom{n}{k} \sqrt{3}^{n-k} (1)^{k}[/imath] I can write this as [imath](\sqrt{3} + 1)^{n}[/imath] but the problem is k=2 and not 0 so i have 2 questions:
1) Will this equal [imath](\sqrt{3} + 1)^{n}[/imath] if k does not equal 0 in our case k=2
2) if not then how i can write this in [imath](x+y)^n[/imath] form when k is not 0 ?

 

[blamocur]

Just subtract the first couple of terms, i.e., the ones corresponding to k=0,1. It won't look like "pure" binomial form, but will look nicer, IMHO, then that clumsy sum.