Chapter 9.1, Problem 68E

To determine

To find the sum:

  ${\displaystyle \sum _{j=3}^5\frac{1}{j^2-3}}$

Answer to Problem 68E

  ${\displaystyle \sum _{j=3}^5\frac{1}{j^2-3}}=\frac{124}{429}$

Explanation of Solution

Given:

  ${\displaystyle \sum _{j=3}^5\frac{1}{j^2-3}}$

Concept Used:

  ${\displaystyle \sum _{i=0}^{n}f\left(x_i\right)}=f\left(x_0\right)+f\left(x_1\right)+\mathrm{......}+f\left(x_n\right)$

Calculation:

In order to determine the sum:

  ${\displaystyle \sum _{j=3}^5\frac{1}{j^2-3}}$

Expand this summation and evaluate.

As,

  $\begin{array}{c}{\displaystyle \sum _{j=3}^5\frac{1}{j^2-3}}=\frac{1}{{\left(3\right)}^2-3}+\frac{1}{{\left(4\right)}^2-3}+\frac{1}{{\left(5\right)}^2-3}\\ =\frac{1}{9-3}+\frac{1}{16-3}+\frac{1}{25-3}\\ =\frac{1}{6}+\frac{1}{13}+\frac{1}{22}\\ =\frac{13+6}{78}+\frac{1}{22}\\ =\frac{19}{78}+\frac{1}{22}\\ =\frac{418+78}{1716}\\ =\frac{496}{1716}\\ =\frac{124}{429}\end{array}$

Thus,

  ${\displaystyle \sum _{j=3}^5\frac{1}{j^2-3}}=\frac{124}{429}$