# How do you use sigma notation to write the sum for 1-1/2+1/4-1/8+...-1/128?

Mar 9, 2017

${\sum}_{k = 0}^{7} {\left(- 1\right)}^{k} \frac{1}{2} ^ k$

#### Explanation:

$1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \ldots - \frac{1}{128}$

For the basic pattern, we note this:

$= \frac{1}{2} ^ 0 - \frac{1}{2} ^ 1 + \frac{1}{2} ^ 2 - \frac{1}{2} ^ 3 + \ldots - \frac{1}{2} ^ k$

In order to work out $k$, we know that:

${2}^{k} = 128 \implies k = 7$

$= \frac{1}{2} ^ 0 - \frac{1}{2} ^ 1 + \frac{1}{2} ^ 2 - \frac{1}{2} ^ 3 + \ldots - \frac{1}{2} ^ 7$

Next, the series is alternating from positive to negative. So we can this:

$= {\left(- 1\right)}^{0} \frac{1}{2} ^ 0 + {\left(- 1\right)}^{1} \frac{1}{2} ^ 1 + {\left(- 1\right)}^{2} \frac{1}{2} ^ 2 + {\left(- 1\right)}^{3} \frac{1}{2} ^ 3 + \ldots + {\left(- 1\right)}^{7} \frac{1}{{2}^{7}}$

The sum is therefore:

${\sum}_{k = 0}^{7} {\left(- 1\right)}^{k} \frac{1}{2} ^ k$