So you have [n + 2] + [(n-1) + 2] + [(n-2) + 2] + ... ... + [1+ 2] + [0+ 2] and want to know how that is equal to n^2/2+ 5n/2- 3? Well, it's NOT! If n= 2, that would be [2+ 2]+ [1+ 2]+ [0+ 2]= 4+ 3+ 2= 9. That is NOT 2^2/2+ 5(2)/2- 3= 2+ 5- 3= 4.

To sum [n + 2] + [(n-1) + 2] + [(n-2) + 2] + ... ... + [1+ 2] + [0+ 2], note that you have n+ 1 terms (from n down to 0), so you are adding 2 n+ 1 times and that sums to 2(n+1)= 2n+ 1. The sum 1+ 2+ ...+ (n-1)+ n is, as you say, an arithmetic progression and sums to n(n+1)/2. So your sum is 2n+1+ n(n+1)/2= 2n+ 1+ n^2/2+ n/2= n^2/2+ 3n/2+ 1.