Evaluate the sum 6/(3^2 - 1) + 6/(4^2 - 1) + 6/(5^2 - 1) + 6/(6^2 - 1) + 6/(7^2 - 1) + ...
+128475 Note that we can write 1/ [ n^2 -1] as 1/[ 2(n -1)] - 1/[2(n + 1)]
So we have
6 ( 1 / [ 2(3-1)] - 1/[2(3 + 1)] + 1/[2 (4-1)] - 1 /[2(4 + 1)] + 1/[2(5-1)] - 1/[2(5 + 1) ] + .......... ) =
6 [ 1/4 - 1/8 + 1/6 - 1/10 + 1/8 - 1/12 + 1/10 - 1/14 + 1/12 - 1/16 + 1/14 - 1/18 + ......]
All the terms in red will "cancel" and we are left with
6 [ 1/4 + 1/6] =
6 [ 10 / 24 ] =
6[ 5/12] =
5 / 2
