Question

Find each sum

$\sum _{n=1}^{80}(2n-5)$

Short answer

The solution is$S_{80}=6080$

Step-by-step solution

Given information

$\sum _{n=1}^{80}(2n-5)$

Calculation

We have, $\sum _{n=1}^{80}(2n-5)$

The series goes $-3,-1,1,....,155$

So, $a_1=-3,d=2,a_n=155$

Now use formula,

$$\begin{gathered} a_n=a_1+(n-1)d \\ \Rightarrow 155=-3+(n-1)2 \\ \Rightarrow 155=-3+2n-2 \\ \Rightarrow 2n=160 \\ \Rightarrow n=80 \end{gathered}$$

To find the sum $S_{80}$

$$\begin{gathered} S_{80}=\frac{80}{2}(-3+155) \\ =6080 \end{gathered}$$