Answer
$$d_1=1 \\d_2=-\dfrac{2}{3} \\d_3=\dfrac{3}{5} \\ d_4=-\dfrac{4}{7} \\ d_5=\dfrac{5}{9}$$
Work Step by Step
We are given that $\{d_n\}=\left\{(-1)^{n+1}\frac{n}{2n-1}\right\}$
In order to determine the first five terms, we will have to substitute $n=1,2,3,4,5$ into the given sequence {$a_n$}:
$$\displaystyle d_1=(-1)^{1+1} \times \frac{1}{2(1)-1}=(1)(1)=1 \\d_2=(-1)^{2+1} \times \frac{2}{2(2)-1}=(-1) \times \frac{2}{3}=-\frac{2}{3} \\d_3=(-1)^{3+1} \times \frac{3}{2(3)-1}=\frac{3}{5} \\ d_4=(-1)^{4+1} \times \frac{4}{2(4)-1}=(-1) \times \frac{4}{7}=-\frac{4}{7} \\ d_5=(-1)^{5+1} \times \frac{5}{2(5)-1}=\frac{5}{9}$$
