Sum of a series

$\frac{1}{8} + \frac{1}{24} + ................ + \frac{1}{2400}$

$\frac{1}{2.4} + \frac{1}{4.6} + \frac{1}{6.8} + .................. + \frac{1}{48.50}$

$\frac{1}{2}( \frac{2}{2.4} + \frac{2}{4.6} + .................. + \frac{2}{48.50})$

$\frac{1}{2}(\frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + .... + \frac{1}{48} -\frac{1}{50})$

$= \frac{1}{2}(\frac{1}{2} - \frac{1}{50})$

$= \frac{24}{100}$

$\frac{1}{8} + \frac{1}{24} + \frac{1}{48} + \frac{1}{80}+ \cdot\cdot\cdot +\frac{1}{2400}$

$= \frac{1}{8} [1+ \frac{1}{3} + \frac{1}{6} + \frac{1}{10}+\cdot\cdot\cdot+\frac{1}{300}]$

$= \frac{1}{8} \sum_{n=1}^{24} [\frac{2}{n(n+1)} ]$

$= \frac{1}{8} \sum_{n=1}^{24} [\frac{2}{n} - \frac{2}{(n+1)} ]$

$= \frac{1}{4} \sum_{n=1}^{24} [\frac{1}{n} - \frac{1}{(n+1)} ]$

The rest is just .......................

First, Lets find out the number of terms in the series. For this purpose only, lets take only denominators and divide them by 8. We get 1,3,6,10,............,300. This is a well known series whose nth term is sum of first n natural numbers or $\frac{n(n+1)}{2}$ . If we solve 300= $\frac{n(n+1)}{2}$ , we get 300 is the 24th term.

Now, to the main solution, we can express the 1st term as $\frac {1}{4} -\frac {1}{8}$ , 2nd term as $\frac {1}{8} -\frac {1}{12}$ , 3rd term as $\frac {1}{12} -\frac {1}{16}$ , 4th term as $\frac {1}{16} -\frac {1}{20}$ . Similarly, the nth term is $\frac {1}{4n} -\frac {1}{4(n+1)}$ . So the last term or the 24th can be expressed as $\frac {1}{96} -\frac {1}{100}$ . If we add all these terms, the negative part of a term and positive part of the next term will get cancelled. Thus only $\frac{1}{4}$ and $-\frac{1}{100}$ will remain. So the sum of the given series is, $\frac { 1 }{ 4 } -\frac { 1 }{ 100 }$ = $\frac{24}{100}$ = 0.24