Question

In Problems 37–54, find each sum.

$\sum _{n=1}^{80}\left(\frac{1}{3}n+\frac{1}{2}\right)$

Short answer

$\sum _{n=1}^{80}\left(\frac{1}{3}n+\frac{1}{2}\right)=1120$

Step-by-step solution

Given Information 

$\sum _{n=1}^{80}\left(\frac{1}{3}n+\frac{1}{2}\right)$

find out first term 

$$\begin{gathered} \sum _{n=1}^{80}\left(\frac{1}{3}n+\frac{1}{2}\right) \\ totalof80terms \\ {a}_n=\frac{1}{3}n+\frac{1}{2} \\ tofindfirstterm,replacen=1 \\ {a}_1=\frac{1}{3}\left(1\right)+\frac{1}{2}=\frac{5}{6} \end{gathered}$$

last term 

$$\begin{gathered} a_n=\frac{1}{3}n+\frac{1}{2} \\ n=80 \\ {a}_{80}=\frac{1}{3}\left(80\right)+\frac{1}{2} \\ {a}_{80}=\frac{160}{6}+\frac{3}{6}=\frac{163}{6} \end{gathered}$$

Sum 

apply sum formula

$$\begin{gathered} S_n=\frac{n}{2}(a_1+a_{80}) \\ {S}_{80}=\frac{80}{2}(a_1+a_{100}) \\ {S}_{80}=\frac{80}{2}\left(\frac{5}{6}+\frac{163}{6}\right) \\ {S}_{80}=40\left(28\right) \\ {S}_{80}=1120 \end{gathered}$$