- #1

- 13

- 0

**Homework Statement**

Prove by mathematical induction:

[itex]\sum^n_{i=1} 1/((2i-1)(2i+1)) = n/(2n+1)[/itex]

**The attempt at a solution**

[itex]\sum^k+1_{i=1} = 1/((2k-1)(2k+1)) + 1/((2(k+1)-1)(2(k+1)+1))[/itex]

= [itex]k/(2k+1) + 1/(4(k+1)^2 + 2(k+1) - 2(k+1) -1)[/itex]

= [itex]k/(2k+1) + 1/(4(k+1)^2 -1)[/itex]

= [itex]k/(2k+1) + 1/(4(k^2 + 2k + 1))[/itex]

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

Solved (thanks Dick):

= [itex]k/(2k+1) + 1/((2k+3)(2k+1))[/itex]

= [itex]k(2k+3)/(2k+3)(2k+1) + 1/((2k+3)(2k+1))[/itex]

= [itex](k(2k+3) + 1)/(2k+3)(2k+1)[/itex]

= [itex](2k^2 + 3k +1)/(2k+3)(2k+1)[/itex]

= [itex]((k+1)(2k+1))/(2k+3)(2k+1)[/itex]

= [itex](k+1)/(2k+3)[/itex]

= [itex](k+1)/(2k+2+1)[/itex]

= [itex](k+1)/(2(k+1)+1)[/itex]

Last edited: