1.17

Step 1: Show that the statement holds for the basis case [math]\displaystyle{ n = 1 }[/math]

[math]\displaystyle{ \frac {1}{i(i+1)} = \frac {n}{n+1} }[/math]

[math]\displaystyle{ \frac {1}{1(1+1)} = \frac {1}{1+1} }[/math]

[math]\displaystyle{ \frac {1}{2} = \frac {1}{2} }[/math]

Since [math]\displaystyle{ 1/2 = 1/2 }[/math], 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.

[math]\displaystyle{ \sum_{i = 1}^{n+1} = \frac{n+1}{n+1+1} = \frac{n}{n+1} + \frac{1}{(n+1)(n+1+1)} }[/math]
[math]\displaystyle{ \frac{n+1}{n+2} = \frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)} }[/math]
[math]\displaystyle{ \frac{n+1}{n+2} = \frac{n(n+2)+1}{(n+1)(n+2)} }[/math]
[math]\displaystyle{ \frac{n+1}{n+2} = \frac{n^2+2n+1}{(n+1)(n+2)} }[/math]
[math]\displaystyle{ \frac{n+1}{n+2} = \frac{(n+1)(n+1)}{(n+1)(n+2)} }[/math]
[math]\displaystyle{ \frac{n+1}{n+2} = \frac{(n+1)}{(n+2)} }[/math]

QED

Back to Chapter 1.