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I know this is a cliche, but you should really show your work (also in the other thread), not just because it makes others happy, but mostly because it is not useful, nor fun, to be spoon-fed an induction proof.

You are probably aware of the general structure of this proof technique, but in what is perhaps its most elementary form it looks like this. You want to prove a statement for all $n \ge n_0$ where $n_0$ is an integer:

*step 0*. You verify the base case, i.e. you verify the statement holds for $n = n_0$. Often, but not always, $n_0 = 0$ or $n_0 = 1$.

*step 1*. You verify: **If** the statement holds for all $n = n_0,\ldots,m$ where $m \ge n_0$ is a certain integer, **then** the statement also holds for $n = m + 1$.

Step 0 cannot be omitted, although it is often tempting to do so.

Now, in these two threads, why don't you start by verifying step 0. If that checks out, then formulate the induction hypotheses. (This hypothesis is the part between**if** and **then** in step 1.) Finally, take all the time that is required to perform the induction step and do not give up on it too easily.

You are probably aware of the general structure of this proof technique, but in what is perhaps its most elementary form it looks like this. You want to prove a statement for all $n \ge n_0$ where $n_0$ is an integer:

Step 0 cannot be omitted, although it is often tempting to do so.

Now, in these two threads, why don't you start by verifying step 0. If that checks out, then formulate the induction hypotheses. (This hypothesis is the part between

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\(\displaystyle \sum_{k=1}^n\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n}{2n+1}\)

I see that in order for the sum on the LHS to go up to \(n+1\), we need to add:

\(\displaystyle \frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac{1}{(2n+1)(2n+3)}\)

And this gives us:

\(\displaystyle \sum_{k=1}^n\left(\frac{1}{(2k-1)(2k+1)}\right)+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\)

Or:

\(\displaystyle \sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\)

Can you show that the RHS is:

\(\displaystyle \frac{n+1}{2(n+1)+1}\) ?

If you can, then you will have derived \(P_{n+1}\) from \(P_n\) thereby completing the proof by induction.

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