First five terms of n^2-3n-6?

Question

need help and/or answers ASAP! cheers

Answer 1

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.