Stuck on last step of Summation of Finite Series question

Question

Find the sum of the following series

$\frac{1}{1\times3} + \frac{7}{3\times5} + \frac {17}{5\times7} + \frac {31}{7\times9} +...+ \frac{2n^2-1}{(2n-1)(2n+1)}$

So, I think I know how to solve this question but my answer doesn’t match the book’s answer.

$ \frac{2n^2-1}{(2n-1)(2n+1)}=\frac{2n^2-1}{4n^2-1}= \frac{1}{2} -\frac{1}{2(2n-1)(2n+1)}$ (Long division)

By partial fraction,

Let $\frac{1}{2(2n-1)(2n+1)} \equiv \frac{A}{2(2n-1)} + \frac{B}{(2n+1)}$

$1\equiv A(2n+1)+2B(2n-1)$

Let $n=-\frac{1}{2}$, B$=-\frac{1}{4}$

Let $n=\frac{1}{2}$, A$=\frac{1}{2}$

$\frac{1}{2(2n-1)(2n+1)} \equiv \frac{1}{4(2n-1)}- \frac{1}{4(2n+1)}$

$$\sum_{r=1}^n \frac{2r^2-1}{(2r-1)(2r+1)} =\sum_{r=1}^n\biggl(\frac{1}{2}-\Bigl(\frac{1}{4(2r-1)}-\frac{1}{4(2r+1)}\Bigr)\biggr)$$

=$\frac{1}{2}- [\frac{1}{4}-\frac{1}{12}$

$+\frac{1}{12}-\frac{1}{20}$

$+\frac{1}{20}-\frac{1}{28}$

.....

$+\frac{1}{4(2n-1)} -\frac{1}{4(2n+1)}]$

$=\frac{1}{2}-[\frac{1}{4}-\frac{1}{4(2n+1)}]$

$=\frac{1}{2}-\frac{2n+1-1}{4(2n+1)}$

$=\frac{1}{2}-\frac{n}{2(2n+1)}$

$=\frac{2n+1-n}{2(2n+1)}$

$=\frac{n+1}{2(2n+1)}$

The answer from the book is $\frac{-n}{2(2n+1)}$

What went wrong?

Answer 1

$\frac{-n}{2(2n+1)}+\frac{1}{2}=\frac{n+1}{2(2n+1)}.$

Hence your solution is correct. In the book they forgot to add $\frac{1}{2}.$