Question

Help with summation notation?

Answer 1

\[\sum_{n=1}^{12} (4n-n^2)\]

Answer 2

wait no wat?

Answer 3

I need to know how to solve that equation with summation notation ^

Answer 4

\[[4(1)-(1)^2] + [4(2) - (2)^2] + ... + [4(12) -(12)^2]\]

Answer 5

becomes -338

Answer 6

oh, ok, i thought so, but i got confused, the first 3 of the series were positive, then they started to go negative. i didn't know if that was possible

Answer 7

You could also do
\[\sum_{n=1}^{12} (4n-n^2)= 4 \sum_{n=1}^{12}n- \sum_{n=1}^{12}n^2\]
Both sums have closed forms. See
http://en.wikipedia.org/wiki/Sum_of_integers
http://en.wikipedia.org/wiki/Square_pyramidal_number

Answer 8

ok thanks