Question

Find the sum.

Answer 1

\[\sum_{n =1}^{12}3n ^{2}+1\]

Answer 2

\[ \sum_{n=1}^{12} 1 = 1 + 1 + 1+ .. +1 \text{ -- 12 times}\]

Answer 3

\[ \sum_{n=1}^{12}3n^2 = 3\sum_{n=1}^{12}n^2 = 3 \times {1 \over 6}12(12+1)(12+1)\]
http://www.math.com/tables/expansion/power.htm