# TADM2E 1.15

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Revision as of 18:22, 11 September 2014 by Algowikiadmin (Talk | contribs)

**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