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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