# What is the sum of 2/3+8/9+26/27+... to n terms ?

Apr 2, 2017

${\sum}_{k = 1}^{n} {a}_{k} = n - \frac{1}{2} + \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}$

#### Explanation:

The general term of a geometric series with initial term $a$ and common ratio $r$ can be written:

${a}_{k} = a \cdot {r}^{k - 1}$

The sum to $n$ terms is given (see footnote) by the formula:

${\sum}_{k = 1}^{n} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$

Note that:

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

$\frac{8}{9} = 1 - \frac{1}{9}$

$\frac{26}{27} = 1 - \frac{1}{27}$

So it seems that the general term of the given series is:

${a}_{k} = 1 - \frac{1}{3} ^ k$

We have:

${\sum}_{k = 1}^{n} \frac{1}{3} ^ k = \frac{\frac{1}{3} \left(1 - {\left(\frac{1}{3}\right)}^{n}\right)}{1 - \frac{1}{3}}$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} \frac{1}{3} ^ k} = \frac{1 - {\left(\frac{1}{3}\right)}^{n}}{2}$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} \frac{1}{3} ^ k} = \frac{1}{2} - \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}$

So:

${\sum}_{k = 1}^{n} {a}_{k} = {\sum}_{k = 1}^{n} \left(1 - \frac{1}{3} ^ k\right)$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} {a}_{k}} = {\sum}_{k = 1}^{n} 1 - {\sum}_{k = 1}^{n} \frac{1}{3} ^ k$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} {a}_{k}} = n - \left(\frac{1}{2} - \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}\right)$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} {a}_{k}} = n - \frac{1}{2} + \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}$

$\textcolor{w h i t e}{}$
Footnote

Given:

${a}_{k} = a {r}^{k - 1}$

We find:

$\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k} = \left(1 - r\right) {\sum}_{k = 1}^{n} a {r}^{k - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = {\sum}_{k = 1}^{n} a {r}^{k - 1} - r {\sum}_{k = 1}^{n} a {r}^{k - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = {\sum}_{k = 1}^{n} a {r}^{k - 1} - {\sum}_{k = 2}^{n + 1} a {r}^{k - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{k = 2}^{n} a {r}^{k - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{k = 2}^{n} a {r}^{k - 1}}}} - a {r}^{n}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = a \left(1 - {r}^{n}\right)$

So dividing both ends by $\left(1 - r\right)$ we get:

${\sum}_{k = 1}^{n} {a}_{k} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$