Chapter F, Problem 5E

To determine

To write: The sum ${\displaystyle \sum _{i=1}^5\sqrt{i}}$ is in expanded form.

Answer to Problem 5E

The expanded form of the sum ${\displaystyle \sum _{k=0}^4\frac{2k-1}{2k+1}}$ is $-1+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9}$.

Explanation of Solution

Definition used:

If $a_m,a_{m+1},\mathrm{...},a_n$ are real numbers and m and n are integers such that $m\le n$, then ${\displaystyle \sum _{i=m}^n{a}_i=a_m+}{a}_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_n$.

Calculation:

By the above definition, the expression ${\displaystyle \sum _{k=0}^4\frac{2k-1}{2k+1}}$ simplified as follows.

$\begin{array}{c}{\displaystyle \sum _{k=0}^4\frac{2k-1}{2k+1}}=\frac{2\left(0\right)-1}{2\left(0\right)+1}+\frac{2\left(1\right)-1}{2\left(1\right)+1}+\frac{2\left(2\right)-1}{2\left(2\right)+1}+\frac{2\left(3\right)-1}{2\left(3\right)+1}+\frac{2\left(4\right)-1}{2\left(4\right)+1}\\ =\frac{-1}{1}+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9}\\ =-1+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9}\end{array}$

Thus, the expanded form of the sum ${\displaystyle \sum _{k=0}^4\frac{2k-1}{2k+1}}$ is $-1+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9}$.