# What is the sum of the first 10 terms given 1, 1/3, 1/9,1/27..?

\color{red}{3/2(1-(1/3)^{10})\approx1.499974597

#### Explanation:

Given series $1 , \frac{1}{3} , \frac{1}{9} , \frac{1}{27} , \setminus \ldots$ is a G.P. with the first term $a = 1$ & common ratio $r = \setminus \frac{\frac{1}{3}}{1} = \setminus \frac{\frac{1}{9}}{\frac{1}{3}} = \setminus \frac{\frac{1}{27}}{\frac{1}{9}} = \setminus \ldots = \frac{1}{3}$

Now, the sum of given series up to $n = 10$ terms

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

$= \setminus \frac{1 \left({\left(\frac{1}{3}\right)}^{10} - 1\right)}{\frac{1}{3} - 1}$

$= \setminus \frac{{\left(\frac{1}{3}\right)}^{10} - 1}{- \frac{2}{3}}$

$= \frac{3}{2} \left(1 - {\left(\frac{1}{3}\right)}^{10}\right)$

$\approx 1.499974597$

Jul 7, 2018

${S}_{10} = \frac{3}{2} \left(\frac{{3}^{10} - 1}{3} ^ 10\right) \approx 1.4999$

#### Explanation:

Here,

$1 , \frac{1}{3} , \frac{1}{9} , \frac{1}{27} , \ldots ,$

The first term $= {a}_{1} = 1$

The common ratio $= r = \frac{\frac{1}{3}}{1} = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{\frac{1}{27}}{\frac{1}{9}} = \frac{1}{3}$

So, the given sequence is Geometric sequence with

color(blue)(a_1=1 and r=1/3

The sum of first $n - t e r m s$ of geometric sequence is :

$\textcolor{b l u e}{{S}_{n} = \frac{{a}_{1} \left(1 - {r}^{n}\right)}{1 - r}} \mathmr{and}$ we have color(blue)(n=10

So,

${S}_{10} = \frac{1 \left(1 - {\left(\frac{1}{3}\right)}^{10}\right)}{1 - \frac{1}{3}}$

${S}_{10} = \frac{\left(1 - \frac{1}{3} ^ 10\right)}{\frac{2}{3}}$

${S}_{10} = \frac{3}{2} \left(\frac{{3}^{10} - 1}{3} ^ 10\right)$

${S}_{10} = \frac{3}{2} \left(\frac{59048}{59049}\right)$

${S}_{10} \approx 1.4999$