Chapter 5.3, Problem 1E

To determine

The sum.

Answer to Problem 1E

  ${\displaystyle \sum _{j=1}^4(j^2+1)}=34$

Explanation of Solution

Given information:

The sum ${\displaystyle \sum _{j=1}^4(j^2+1)}$

Formulae used:

  $\begin{array}{l}{\displaystyle \sum _{j=1}^n{j}^2}=\frac{n(n+1)(2n+1)}{6}\\ {\displaystyle \sum _{j=1}^{n}1=n}\end{array}$

Calculation:

Consider ${\displaystyle \sum _{j=1}^4(j^2+1)}$

  $\begin{array}{l}{\displaystyle \sum _{j=1}^4(j^2+1)}={\displaystyle \sum _{j=1}^4{j}^2+{\displaystyle \sum _{j=1}^{4}1}}\\ {\displaystyle \sum _{j=1}^4(j^2+1)}=\frac{4(4+1)(2(4)+1)}{6}+4\\ {\displaystyle \sum _{j=1}^4(j^2+1)}=30+4\\ {\displaystyle \sum _{j=1}^4(j^2+1)}=34\end{array}$

Therefore ${\displaystyle \sum _{j=1}^4(j^2+1)}=34$