Lets rewrite the summation:
Summation from 1 to 12 of (2n+1) is equal to the summation of 1 to 12 of (2n) plus the summation from 1 to 12 of (1).
The summation from 1 to any number ([tex]n[/tex]) is given by:
[tex] \frac{n(n+1)}{2} [/tex]
So, the first term of our rewritten summation is equal to [tex]n(n+1)[/tex].
The summation from 1 to any number ([tex]n[/tex]) of a constant (lets call it [tex]c[/tex]) is given by:
[tex]nc[/tex]
So, the whole summation can be written as:
[tex]n(n+1)+nc=n(n+1+c)[/tex]
For your problem we have [tex]n=12[/tex] and [tex]c=1[/tex].
So, the answer is 168.
