How do i solve this trigonometry question which involves Sigma?

Barron's Regents book is not as specific in explaining how to solve several problems.

The Sigma's bottom notation is (i=1), notation above is 8. So it's Sigma E, above is 8, bottom is i = 1 and then (3i - i^2).

I know it's -96, but have no clue how it came to be -96.



ANSWER:
Here's how you do it step by step. :)

Sigma always denotes SUM. Therefore, you have to add all the values of the expression by substituting i = 1 to 8.

For,

i = 1: 3(1) - (1)^2 = 3 -1 = 2
i = 2: 3(2) - (2)^2 = 6 - 4 = 2
i = 3: 3(3) - (3)^2 = 9 - 9 = 0
i = 4: 3(4) - (4)^2 = 12 - 16 = -4
i = 5: 3(5) - (5)^2 = 15 - 25 = -10
i = 6: 3(6) - (6)^2 = 18 - 36 = -18
i = 7: 3(7) - (7)^2 = 21 - 49 = -28
i = 8: 3(8) - (8)^2 = 24 - 64 = -40

Therefore, the sum is: 2 + 2 + 0 + (-4) + (-10) + (-18) + (-28) + (-40) = -96.

Hope this helps.