Question 804988
1/2 + 3/4 + 5/6 + ... 99/100
<pre>
The denominators are the even natural numbers 2,4,6,8, and the pattern is 2n for n = 1,...,50

The numerators are 1 less than the denominators, 1,3,5,7, so we subtract 1
from that pattern 2n-1,

So the nth term is {{{(2n-1)/(2n)}}}, where n goes from 1 to 50, so we can write that as
 
{{{sum(((2n-1)/(2n)),n=1,50)}}}

We can simplify, if desired:

{{{(2n-1)/(2n)}}} = {{{(2n)/(2n)}}}{{{""-""}}}{{{1/(2n)}}} = 1 - {{{1/(2n)}}}, so we write the summation as

{{{sum((1-1/(2n)),n=1,50)}}}

and if we like we can distribute the summation as

 {{{sum((1),n=1,50)}}}{{{""-""}}}{{{sum((1/(2n)),n=1,50)}}}

Then the first summation is just the sum of 50 ones, which is just 50.
So we can simplify it to

{{{50}}}{{{""-""}}}{{{sum((1/(2n)),n=1,50)}}}

Edwin</pre>