Chapter 9.2, Problem 60E

To determine

To find the partial sum of “ ${\displaystyle \sum _{n=1}^{75}12n-9}$ ”

Answer to Problem 60E

The partial sum of ${\displaystyle \sum _{n=1}^{75}12n-9}$ is $33525$

Explanation of Solution

Given information :

Given series ${\displaystyle \sum _{n=1}^{75}12n-9}$

As we have ${\displaystyle \sum _{n=1}^{75}12n-9}=3+15+27+\mathrm{39...............}$ ,

For given arithmetic series, $a_1=3$ and $d=15-3=12$

And the $n^{th}$ term is $a_n=12n-9$

Therefore, $a_{75}=12\left(75\right)-9=891$

Thus, the sum of the first terms $75$ is

  $\begin{array}{l}{S}_{75}=\left(\frac{75}{2}\right)\left(a_1+a_{100}\right)\\ =\left(\frac{75}{2}\right)\left(3+891\right)\\ =\left(75\right)\left(447\right)\\ \therefore S_{75}=33525\end{array}$

Hence, the partial sum of ${\displaystyle \sum _{n=1}^{75}12n-9}$ is $33525$