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January 21, 2017
Posted by **aya** on Wednesday, November 2, 2011 at 4:06pm.

- mathematical induction -
**Steve**, Wednesday, November 2, 2011 at 4:52pmFirst, 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,...