Posted by aya on Wednesday, November 2, 2011 at 4:06pm.
First, verify for n=1:
1/1*3 = 1/(3) -- true
Assume it is true for n:
1/1*3 + 1/3*5 + ... + 1/(2n-1)(2n+1) = n/(2n+1)
Now, add the next term to both sides:
1/1*3 + 1/3*5 + ... + 1/(2n-1)(2n+1) + 1/(2n+1)(2n+3) = n/(2n+1) + 1/(2n+1)(2n+3)
= [n(2n+3) + 1]/(2n+1)(2n+3)
= (2n^2 + 3n + 1)/(2n+1)(2n+3)
= (2n+1)(n+1)/(2n+1)(2n+3)
= (n+1)/(2(n+1)+1)
This is the formula, with n replaced by n+1.
So, if it's true for n, it's true for n+1.
It's true for n=1, so therefore also for n=2,3,4,...