If $a_1,a_2,a_3,\mathrm{....},a_{2n+1}$ are in A.P., then

$\frac{a_{2n+1}–a_1}{a_{2n+1}+a_1}+\frac{a_{2n}–a_2}{a_{2n}+a_2}+\mathrm{....}+\frac{a_{n+2}–a_n}{a_{n+2}+a_n}$ is equal to

In an A.P. the sum of the terms equidistant from the beginning and the end is always same and is equal to the sum of first and term. Therefore,

           $a_1+a_{2n+1}=a_2+a_{2n}=\mathrm{....}=a_n+a_{n+2}$

$\therefore$       $\frac{a_{2n+1}–a_1}{a_{2n+1}+a_1}+\frac{a_{2n}–a_2}{a_{2n}+a_2}+\mathrm{....}+\frac{a_{n+2}–a_n}{a_{n+2}+a_n}$

         $=\frac{2nd+\left(2n–2\right)d+\mathrm{...}+2d}{a_{2n+1}+a_1}$

          $=2d\frac{n(n+1)}{2}\times \frac{1}{a_{2n+1}+a_1}$

        $=2d\frac{n(n+1)}{2}\times \frac{1}{2a_{n+1}}[\because a_1+a_{2n+1}=2a_{n+1}]$

        $=d\frac{n(n+1)}{2}\times \frac{1}{a_{n+1}}=\frac{n(n+1)}{2}.\frac{a_2–a_1}{a_{n+1}}[\because d=a_2–a_1]$