# How do you use sigma notation to write the sum for 1/4+3/8+7/16+15/32+31/64?

Feb 19, 2017

$\frac{1}{4} + \frac{3}{8} + \frac{7}{16} + \frac{15}{32} + \frac{31}{64} = {\sum}_{r = 1}^{5} \frac{{2}^{n} - 1}{2} ^ \left(n + 1\right)$

#### Explanation:

If we look at the sequence for the terms in denominator we have:

$\left\{4 , 8 , 16 , 32 , 64\right\}$

It should be obvious that these numbers are successive powers of $2$, with the first term $2$ missing so the general term of the above sequence would be:

${u}_{n} = {2}^{n + 1}$ where $n \in \left\{1 , 2 , 3 , 4 , 5\right\}$

And now we look at the sequence for the terms in the numerator

$\left\{1 , 3 , 7 , 15 , 31\right\}$

and note that the terms are one less than successive powers of $2$, so the general term of this sequence be:

${u}_{n} = {2}^{n} - 1$ where $n \in \left\{1 , 2 , 3 , 4 , 5\right\}$

Hence a general term of the series would be:

${u}_{n} = \frac{{2}^{n} - 1}{2} ^ \left(n + 1\right)$ where $n \in \left\{1 , 2 , 3 , 4 , 5\right\}$

And hence we can write the finite series using sigma notation as

$\frac{1}{4} + \frac{3}{8} + \frac{7}{16} + \frac{15}{32} + \frac{31}{64} = {\sum}_{r = 1}^{5} \frac{{2}^{n} - 1}{2} ^ \left(n + 1\right)$

NB: The sum evaluates to $\frac{129}{64}$

Feb 19, 2017

${\sum}_{n = 1}^{5} \frac{{2}^{n} - 1}{{2}^{n + 1}}$

#### Explanation:

Note that:

$\frac{1}{4} = \frac{{2}^{1} - 1}{2} ^ 2$

$\frac{3}{8} = \frac{{2}^{2} - 1}{2} ^ 3$

etc...

so the sum is:

${\sum}_{n = 1}^{5} \frac{{2}^{n} - 1}{{2}^{n + 1}}$

We can also write it as:

$S = {\sum}_{n = 1}^{5} \frac{{2}^{n} - 1}{{2}^{n + 1}} = {\sum}_{n = 1}^{5} \left(\frac{1}{2} - \frac{1}{{2}^{n + 1}}\right) = \frac{5}{2} - \frac{1}{4} {\sum}_{n = 0}^{4} {\left(\frac{1}{2}\right)}^{n}$

This last expression allows us also to calculate the sum, as we have the partial sum of a geometric series:

${\sum}_{k = 0}^{n} {x}^{k} = \frac{1 - {x}^{k + 1}}{1 - x}$

$S = \frac{5}{2} - \frac{1}{4} {\sum}_{n = 0}^{4} \frac{1}{2} ^ n = \frac{5}{2} - \frac{1}{4} \frac{1 - \frac{1}{2} ^ 5}{1 - \frac{1}{2}} = \frac{5}{2} - \frac{1}{4} \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{5}{2} - \frac{31}{64} = \frac{129}{64}$