# For the sequence 1/3, 1/3^2 ,1/3^3 ,1/3^4 ,1/3^5,…, ?

## its fifth partial sum S5= its sixth partial sum S6=

Jun 21, 2016

${S}_{5} = \frac{121}{243.}$
${S}_{6} = \frac{364}{729.}$

#### Explanation:

For the Geometric seq. $a , a r , a {r}^{2} , a {r}^{3} , \ldots .$, the sum ${S}_{n}$ of first $n$ terms is given by ${S}_{n} = \frac{a \cdot \left(1 - {r}^{n}\right)}{1 - r} .$

Here, $a = \frac{1}{3} , r = \frac{1}{3} ,$ so, ${S}_{n} = \frac{\left(\frac{1}{3}\right) \left\{1 - {\left(\frac{1}{3}\right)}^{n}\right\}}{1 - \frac{1}{3}} ,$ i.e., $\frac{1}{2} \left\{1 - {\left(\frac{1}{3}\right)}^{n}\right\} .$

Accordingly, ${S}_{5} = \frac{1}{2} \left\{1 - {\left(\frac{1}{3}\right)}^{5}\right\} = \frac{1}{2} \left(1 - \frac{1}{243}\right) = \frac{242}{2 \cdot 243} = \frac{121}{243.}$

For ${S}_{6}$, we use a little trick, by observing that, ${S}_{6} = {S}_{5} + {t}_{6} ,$ [Here, ${t}_{6}$=sixth term]
=$\frac{121}{243} + \frac{1}{3} ^ 6 = \frac{121}{243} + \frac{1}{243 \cdot 3} = \frac{121 \cdot 3 + 1}{243 \cdot 3} = \frac{364}{729}$