Answer
The first five terms are:
$d_1 = 1;
\\d_2= -\frac{2}{3};
\\d_3= \frac{3}{5};
\\d_4= -\frac{4}{7};
\\d_5= \frac{5}{9}$
Work Step by Step
Substitute $n=1, 2, 3, 4, 5$ to obtain:
$d_1 = (-1)^{1-1}\left(\dfrac{1}{2(1)-1}\right)=(-1)^0(\frac{1}{1})=1(1)=1;
\\d_2= (-1)^{2-1}\left(\dfrac{2}{2(2)-1}\right)=(-1)^1(\frac{2}{3})=-1(\frac{2}{3})=-\frac{2}{3};
\\d_3= (-1)^{3-1}\left(\dfrac{3}{2(3)-1}\right)=(-1)^2(\frac{3}{5})=1(\frac{3}{5})=\frac{3}{5};
\\d_4= (-1)^{4-1}\left(\dfrac{4}{2(4)-1}\right)=(-1)^3(\frac{4}{7})=-1(\frac{4}{7})=-\frac{4}{7};
\\d_5= (-1)^{5-1}\left(\dfrac{5}{2(5)-1}\right)=(-1)^4(\frac{5}{9})=1(\frac{5}{9})=\frac{5}{9}$
