induction question

indefiniteintegral

New member
Joined
Mar 8, 2020
Messages
5
Hi so I have this question for an intro to proofs class (I've never taken proofs)

I attached a picture of my rough work it's kind of messy but if you can't read it I am happy to re write everything much clearer.

First off I just want to say this is just rough work I would never hand in a proof that looks like this.

So in terms of the proof, simply, I am trying to do 3 things

1) show the basis
2) n=k
3)n=k+1
and at the end of 3) the left side should = the right side, thus completing the proof

I have those (1,2,3) marked in the picture.

I do not want the answer!
I just would like to know where I messed up.

It has to be (I think) in step 3) where I am doing n = k + 1

My algebra is wrong there somewhere, if someone could kindly point out where I messed up I would appreciate it.
 

Attachments

  • isuck.jpg
    isuck.jpg
    1.3 MB · Views: 8
It appears the induction hypothesis \(P_n\) is:

[MATH]\sum_{k=1}^{n}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n}{2n+1}[/MATH]
We can see the base case \(P_1\) is true, and so for our induction step, we could write:

[MATH]\sum_{k=1}^{n}\left(\frac{1}{(2k-1)(2k+1)}\right)+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac{n}{2n+1}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}[/MATH]
[MATH]\sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}[/MATH]
[MATH]\sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n(2n+3)+1}{(2n+1)(2n+3)}[/MATH]
This is where you are, and we agree. To continue:

[MATH]\sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{2n^2+3n+1}{(2n+1)(2n+3)}[/MATH]
[MATH]\sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{(2n+1)(n+1)}{(2n+1)(2n+3)}[/MATH]
[MATH]\sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n+1}{2n+3}[/MATH]
[MATH]\sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n+1}{2(n+1)+1}[/MATH]
We have now derived \(P_{n+1}\) from \(P_n\) thereby completing the proof by induction. You were almost there, you just neext to expand and factor your numerator.
 
Thank you so much, its funny because on the top right,
I actually did expand and factor it, but I thought it was wrong so I crossed it out!! I was just so tired my brain wasn't working I think.

greatly appreciated :)
 
Top