Chapter 8.1, Problem 143E

To determine

To write: the first five terms of the sequence and an expression for nth partial sum.

Answer to Problem 143E

The first five terms are

  $\frac{1}{4},\frac{1}{12},\frac{1}{24},\frac{1}{40},\frac{1}{60}$

The expression for nth partial sum is $\frac{1}{2}-\frac{1}{2n+2}$ .

Explanation of Solution

Given information:

Given sequence

  $a_n=\frac{1}{2n}-\frac{1}{2n+2}$

Calculation:

Let, $a_i$ is the $i^{th}$ term of the sequence then,

The first five terms given by the sequence $a_n=\frac{1}{2n}-\frac{1}{2n+2}$ are,

  $\begin{array}{l}{a}_1=\frac{1}{2\times 1}-\frac{1}{2\times 1+2}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\\ a_2=\frac{1}{2\times 2}-\frac{1}{2\times 2+2}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\\ a_3=\frac{1}{2\times 3}-\frac{1}{2\times 3+2}=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}\\ a_4=\frac{1}{2\times 4}-\frac{1}{2\times 4+2}=\frac{1}{8}-\frac{1}{10}=\frac{1}{40}\\ a_5=\frac{1}{2\times 5}-\frac{1}{2\times 5+2}=\frac{1}{10}-\frac{1}{12}=\frac{1}{60}\end{array}$

Hence, the first five terms are

  $\frac{1}{4},\frac{1}{12},\frac{1}{24},\frac{1}{40},\frac{1}{60}$

Now, add all terms up to nth term for expression for nth partial sum

  $\begin{array}{c}{\displaystyle \sum a_n}=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\mathrm{...}+a_n\\ =\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\mathrm{...}+\frac{1}{2n-2}-\frac{1}{2n}+\frac{1}{2n}-\frac{1}{2n+2}\\ =\frac{1}{2}-\frac{1}{2n+2}\end{array}$

Hence, expression for nth partial sum is $\frac{1}{2}-\frac{1}{2n+2}$ .