Calculate the sum of this series?

I assume that you meant to type i^2 instead of t^2.

sum(i = -3 to n)(n^2 + i^2 + 2) =

sum(i = -3 to n)(n^2 + 2) + sum(i = -3 to n)(i^2) =

(n^2 + 2)*sum(i = -3 to n)(1) + sum(i = -3 to 0)(i^2) + sum(i = 1 to n)(i^2) =

(n^2 + 2)*(n - (-3) + 1) + ((-3)^2 + (-2)^2 + (-1)^2 + 0^2) + [n*(n + 1)*(2n + 1) / 6] =

(n^2 + 2)*(n + 4) + (9 + 4 + 1) + (1/6)*(2*n^3 + 3*n^2 + n) =

(1/6)*(6*(n^3 + 4*n^2 + 2n + 8) + 6*(14) + 2*n^3 + 3*n^2 + n) =

(1/6)*(8*n^3 + 27*n^2 + 13n + 132).

Edit: O.k., that is an odd question then. So the answer would simply be

(n^2 + t^2 + 2)*sum(i = -3 to n)(1) =

(n^2 + t^2 + 2)*(n - (-3) + 1) = (n^2 + t^2 + 2)*(n + 4), as the other answerer has.

I guess the question is more about what you do when the index starts

at a number other than 1. So as you probably already know, if the index starts

at j and goes up to k, with k > j, then there are (k - j + 1) terms in the sum.

The question would be much better if the t were an i. :)

Σ(i=-3,n) (n^2+t^2+2) = (n+4) (n^2+t^2+2)

ℬ ℋ