Chapter 13, Problem 1RE

Calculate ${\displaystyle \underset{k=1}{\overset{8}{\sum }}\left(k^2+1\right)}$.

To determine

To calculate: The value of the expression is ${\displaystyle \sum _{k=1}^8\left(k^2+1\right)}$.

Answer to Problem 1RE

Solution:

The value of the expression ${\displaystyle \sum _{k=1}^8\left(k^2+1\right)}$ is $212$.

Explanation of Solution

Given Information:

The provided expression is ${\displaystyle \sum _{k=1}^8\left(k^2+1\right)}$.

Formula Used:

The sum of n numbers $x_1,x_2,x_3,\dots ,x_n$ is given by

${\displaystyle \sum _{k=1}^n{x}_k}=x_1+x_2+\dots +x_n$

The value of the sum ${\displaystyle \sum _{k=1}^{n}1}$ is n.

${\displaystyle \sum _{k=1}^{n}1}=n$

The sum of the square of first n natural numbers,

${\displaystyle \sum _{k=1}^n{k}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Calculation:

Consider the sum ${\displaystyle \sum _{k=1}^8\left(k^2+1\right)}$.

Separate the components of the expression as follows:

${\displaystyle \sum _{k=1}^8\left(k^2+1\right)}={\displaystyle \sum _{k=1}^8\left(k^2\right)}+{\displaystyle \sum _{k=1}^8\left(1\right)}$

Apply the formula for the summation of the expressions,

Firstly, the sum of the first expression is:

$\begin{array}{c}{\displaystyle \sum _{k=1}^n\left(k^2\right)}=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\\ {\displaystyle \sum _{k=1}^8\left(k^2\right)}=\frac{8\left(9\right)\left(17\right)}{6}\\ =204\end{array}$

Secondly, the sum of the second expression is:

$\begin{array}{c}{\displaystyle \sum _{k=1}^n\left(1\right)}=n.1\\ {\displaystyle \sum _{k=1}^8\left(1\right)}=8.1\\ =8\end{array}$

Add both the values for the expressions:

$\begin{array}{c}{\displaystyle \sum _{k=1}^8\left(k^2+1\right)}={\displaystyle \sum _{k=1}^8\left(k^2\right)}+{\displaystyle \sum _{k=1}^8\left(1\right)}\\ =204+8\\ =212\end{array}$

Hence, the value of the expression ${\displaystyle \sum _{k=1}^8\left(k^2+1\right)}$ is $212$.