# How do you use sigma notation to write the sum for 1/(1*3)+1/(2*4)+1/(3*5)+...+1/(10*12)?

Feb 14, 2017

${\sum}_{n = 1}^{12} \frac{1}{n \left(n + 2\right)}$

#### Explanation:

${\sum}_{n = 1}^{12} \frac{1}{n \left(n + 2\right)}$

Feb 14, 2017

I give the sum here. $\frac{1}{2} \left(1 + \frac{1}{2} - \frac{1}{11} - \frac{1}{12}\right) = \frac{175}{264}$

#### Explanation:

The $\sum$ notation has already appeared, in the other answer.

As a matter of interest, I perform summation.

$\frac{1}{n \left(n + 2\right)} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n + 2}\right)$.

So, the sum is

$\frac{1}{2} \left(1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \ldots + \frac{1}{9} - \frac{1}{11} + \frac{1}{10} - \frac{1}{12}\right)$

$= \frac{1}{2} \left(1 + \frac{1}{2} - \frac{1}{11} - \frac{1}{12}\right) = \frac{175}{264}$