Chapter 9.1, Problem 105E

To determine

The error in finding the sum of the series “ ${\displaystyle \sum _{k=1}^4\left(3+2k^2\right)}$ ”

Answer to Problem 105E

  ${\displaystyle \sum _{k=1}^{4}3}=12$ .

Explanation of Solution

Given information :

Given series ${\displaystyle \sum _{k=1}^4\left(3+2k^2\right)}={\displaystyle \sum _{k=1}^{4}3}+{\displaystyle \sum _{k=1}^4\left(3+2k^2\right)}$

  $\begin{array}{l}={\displaystyle \sum _{k=1}^4\left(3+2k^2\right)}\\ =3+{\displaystyle \sum _{k=1}^4\left(2k^2\right)}\\ =3+2+8+18+32\\ =63\end{array}$

The error in finding the sum of the series ${\displaystyle \sum _{k=1}^4\left(3+2k^2\right)}$ is ${\displaystyle \sum _{k=1}^{4}3}\ne 3$

Thus, ${\displaystyle \sum _{k=1}^{4}3}=3+3+3+3=12$

Hence, ${\displaystyle \sum _{k=1}^{4}3}$ has to be $12$