Induction And Proofs Find a formula for $\sum_{i=1}^{n} \frac{1}{(2i-1)(2i+1)}$ and prove that it holds for all $n \geq 1$ I don't know how to solve this particular problem, can someone help me please. Thanks 
 A: Hint: Expand the summand by partial fractions to see that the partial sums telescope: 
$$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\\
\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\sum_{k=1}^n\left(\frac{1}{2(2k-1)}-\frac{1}{2(2k+1)}\right)\\
=\frac{1}{2(2\cdot 1-1)}-\frac{1}{2(2n+1)}$$
A: Here are the steps to take to prove a formula for the sum by induction.  I'll leave it to you to work out the details.
1)  Compute values of the sum for small values of $n$.  Doing so for 5 or more values of $n$ will hopefully reveal a pattern that you can conjecture holds for all $n$.
2)  Now that you have a conjectural formula for the sum, check to make sure that it holds in the base case when $n=1$.  This should be easy after step 1.
3)  Now the key step is that you will assume your conjectured formula is correct for an arbitrary value of $n$.  Under that assumption, prove that the formula holds for $n+1$.  Having done that, you will have proven that if your sum formula holds for a particular value for $n$, then it also holds for all values of $n+1$.
4)  You can now conclude that your conjectured formula holds for $n$ by the Principle of Mathematical Induction (PMI).  Indeed, the PMI says that if a statement is true for $n=1$ (what you prove in step 2) and if it is also the case that whenever a statement holds for a value of $n$, it also holds for the value $n+1$ (what you prove in step 3), then the statement holds for all natural numbers.
