TADM2E 1.15

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Step 1: Show that the statement holds for the basis case $ n = 1 $

$ \frac {1}{i(i+1)} = \frac {n}{n+1} $

$ \frac {1}{1(1+1)} = \frac {1}{1+1} $

$ \frac {1}{2} = \frac {1}{2} $

Since $ 1/2 = 1/2 $, the basis case is true.

Step 2: Assume that that summation is true up to n.

Step 3: Show that on the assumption that the summation is true for n, it follows that it is true for n + 1.

$ \sum_{i = 1}^{n+1} = \frac{n+1}{n+1+1} = \frac{n}{n+1} + \frac{1}{(n+1)(n+1+1)} $
$ \frac{n+1}{n+2} = \frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)} $
$ \frac{n+1}{n+2} = \frac{n(n+2)+1}{(n+1)(n+2)} $
$ \frac{n+1}{n+2} = \frac{n^2+2n+1}{(n+1)(n+2)} $
$ \frac{n+1}{n+2} = \frac{(n+1)(n+1)}{(n+1)(n+2)} $
$ \frac{n+1}{n+2} = \frac{(n+1)}{(n+2)} $


QED