The sum of the first n natural squares is n(n+1)(2n+1)/6.
The sum of the first n natural numbers is n(n+1)/2.
So the sum to n terms of the given series is n(n+1)(2n+1)/6-3n(n+1)/2-6n=n(n²-3n-22)/3.
Plug in n=5: (5)(25-15-22)/3=-20.
Or, you could just work out each term:
-8-8-6-2+4=-20.