Question

In Problems 17–28, write down the first five terms of each sequence.

$\left\{d_n\right\}=\left\{{(-1)}^{n-1}\left(\frac{n}{2n-1}\right)\right\}$

Short answer

The first five terms of the given sequence are$1,-\frac{2}{3},\frac{3}{5},-\frac{4}{7},\frac{5}{9}$

Step-by-step solution

Step 1. Write the given sequence.

The given sequence is$\left\{d_n\right\}=\left\{{(-1)}^{n-1}\left(\frac{n}{2n-1}\right)\right\}$

Step 2. Find the first five terms.

To find the first terms in the sequence we substitute $1,2,3,4and5inton$

$$\begin{gathered} \left\{d_1\right\}={\left(-1\right)}^{1-1}\frac{1}{2·1-1}=1 \\ \left\{d_2\right\}={\left(-1\right)}^{2-1}\frac{2}{2·2-1}=-\frac{2}{3} \\ \left\{d_3\right\}={\left(-1\right)}^{3-1}\frac{3}{2·3-1}=\frac{3}{5} \\ \left\{d_4\right\}={\left(-1\right)}^{4-1}\frac{4}{2·4-1}=-\frac{4}{7} \\ \left\{d_5\right\}={\left(-1\right)}^{5-1}\frac{5}{2·5-1}=\frac{5}{9} \end{gathered}$$