Prove by Induction: \sum^n_{i=1} 1/((2i-1)(2i+1)) = n/(2n+1)

[dystplan]

Homework Statement
Prove by mathematical induction:
\sum^n_{i=1} 1/((2i-1)(2i+1)) = n/(2n+1)

The attempt at a solution
\sum^k+1_{i=1} = 1/((2k-1)(2k+1)) + 1/((2(k+1)-1)(2(k+1)+1))

= k/(2k+1) + 1/(4(k+1)^2 + 2(k+1) - 2(k+1) -1)

= k/(2k+1) + 1/(4(k+1)^2 -1)

= k/(2k+1) + 1/(4(k^2 + 2k + 1))

= k/(2k+1) + 1/(4k^2 8k + 3) - 1)

Solved (thanks Dick):

= k/(2k+1) + 1/((2k+3)(2k+1))

= k(2k+3)/(2k+3)(2k+1) + 1/((2k+3)(2k+1))

= (k(2k+3) + 1)/(2k+3)(2k+1)

= (2k^2 + 3k +1)/(2k+3)(2k+1)

= ((k+1)(2k+1))/(2k+3)(2k+1)

= (k+1)/(2k+3)

= (k+1)/(2k+2+1)

= (k+1)/(2(k+1)+1)

 

[Dick]

You want to show your result is equal to (k+1)/(2*(k+1)+1). You are going to have a hard time doing that because there's a mistake. 1/(4*(k+1)^2-1) is not equal to 1/(4k^2+2k).

 

[dystplan]

You want to show your result is equal to (k+1)/(2*(k+1)+1). You are going to have a hard time doing that because there's a mistake. 1/(4*(k+1)^2-1) is not equal to 1/(4k^2+2k).

Thanks; I've fixed that problem; durh.

Still have no idea how to get from

1/(4k^2+8k+3)−1) (or 1/((2k+3)(2k+1)) ) to (k+1)/(2*(k+1)+1)

 

[Dick]

You want to go from k/(2k+1)+1/((2k+1)*(2k+3)) to (k+1)/(2*(k+1)+1) or (k+1)/(2k+3). It's just algebra. Put everything over a common denominator. Look for common factors to cancel, etc. You don't need to think hard about it before you try just turning the crank and see what happens.

 

[dystplan]

Thanks for the help Dick.