Example (definition to follow)

1 Show that (for any integer $n > 1$):

$$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(n-1)\cdot n} = \frac{n-1}{n}$$

.

OK, you could prove this by breaking each term $\frac{1}{k\cdot k+1}$ down into $\frac{1}{k} - \frac{1}{k+1}$ and turning the left hand side into a telescoping sum. But another way to do this problem is to first show that the formula is true for the smallest possible value for $n$ (that is, when $n=2$, the two sides of the equation are identical) and then to show that, once we have confirmed the formula is ok for some number $n$, we can add $\frac{1}{n \cdot (n+1)}$ to both sides, do a little algebra on the right side, and obtain the formula:

$$\begin{eqnarray*} \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} +... ... (n-1)\cdot(n+1) + 1}{n \cdot (n+1)} \\ &=& \frac{ n}{n+1} \\ \end{eqnarray*}$$

which is exactly the same formula, but applied to the number $n+1$. Putting this together, means that we have shown the equation is true for all positive integers greater than 1. Haven't we?