How to find the sum of the series?

Question

$$\frac{1}{1.3} + \frac{7}{1.3.5} + \frac{17}{1.3.5.7} + \frac{31}{1.3.5.7.9} +...... upto \space 10 \space terms$$

I found the nth term of the equation but don't know how to proceed from there?

$$ T_n = \frac{2n^2 -1 }{1.3.5.7...(2n-1)(2n+1)}$$

Answer 1

In double factorial notation, this is $$\sum^{10}_{n=1}\frac{2n^2-1}{(2n+1)!!}.$$ But $$2n^2-1=\frac{(2n-1)(2n+1)-1}2,$$ and that means $$\frac{2n^2-1}{(2n+1)!!}=\frac12\left(\frac1{(2n-3)!!}-\frac1{(2n+1)!!}\right),$$ i.e. we have a telescoping series, so \begin{align}\sum^{10}_{n=1}\frac{2n^2-1}{(2n+1)!!}&=\frac12\left(\frac1{(-1)!!}+\frac1{1!!}-\frac1{19!!}-\frac1{21!!}\right)=\frac12\left(2-\frac1{654729075}-\frac1{13749310575 }\right)\\&=1-\frac1{1249937325}.\end{align}