Answer
$1120$
Work Step by Step
We have to determine the sum:
$S=\sum_{n=1}^{80} \left(\dfrac{1}{3}n+\dfrac{1}{2}\right)$
$S$ is the sum of an arithmetic sequence.
Determine its first term and the common difference:
$a_1=\dfrac{1}{3}(1)+\dfrac{1}{2}=\dfrac{5}{6}$
$d=\left(\dfrac{1}{3}(k+1)+\dfrac{1}{2}\right)-\left(\dfrac{1}{3}k+\dfrac{1}{2}\right)=\dfrac{1}{3}$
The number of terms is 80, so we have to determine the sum of the first $80$ terms. We use the formula:
$S_n=\dfrac{n(a_1+a_n)}{2}$
$a_1=\dfrac{5}{6}$
$a_{80}=\dfrac{1}{3}(80)+\dfrac{1}{2}=\dfrac{163}{6}$
$\sum_{n=1}^{80} \left(\dfrac{1}{3}n+\dfrac{1}{2}\right)=\dfrac{80\left(\dfrac{5}{6}+\dfrac{163}{6}\right)}{2}=1120$
