But they are fractions

Let $\large S = \frac{1}{2^3} + \frac{1}{2^6} + \frac{1}{2^9} + \ldots$
$\large 2^3\times S = 1 + \frac{1}{2^3} + \frac{1}{2^6} + \frac{1}{2^9} + \ldots$
$\large 8S = 1 + S$
$\large S = \frac{1}{7}$

As we can see that it is a infinite GP Series: $S=\dfrac{1}{2^3}+\dfrac{1}{2^6}+\dfrac{1}{2^9}+\ldots$ Its sum has the formula: $S=\dfrac{a}{1-r}$ Here $a=\frac{1}{8}$ and r= $\frac{1}{8}$ . Therefore: $S=\dfrac{\frac{1}{8}}{1-\frac{1}{8}}$ $S=\dfrac{1}{8-1}$ $\boxed{S=\dfrac{1}{7}}$

Let as assume that
$\frac{1}{2^3} + \frac{1}{2^6} + \frac{1}{2^9} + ... = x$
Taking $\frac{1}{2^3}$ out as a common factor from the left hand side, we get,
$\frac{1}{2^3}[1+\frac{1}{2^3}+\frac{1}{2^6}+\frac{1}{2^9}+...]=x$
$\Rightarrow \frac{1}{2^3}[1+x]=x$
$\Rightarrow 1+x=(2^3)x$
$\Rightarrow 1=8x-x$
$\Rightarrow 7x=1$
$\Rightarrow x= \frac{1}{7}$
$\Rightarrow \frac{1}{2^3}+\frac{1}{2^6}+\frac{1}{2^9}+...=\frac{1}{7}$

I was trying to find out the answer without writing, and this is one of the way that comes up.

Let $x$ is the value of the sequence, one could see that

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

$\Leftrightarrow x = \frac{8}{7} * \frac{1}{8} = \frac{1}{7}$

Thus, the answer is $\frac{1}{7}$

Let $S=\displaystyle\frac{1}{2^3}+\displaystyle\frac{1}{2^6}+\displaystyle\frac{1}{2^9}+...= \displaystyle\sum_{i=0}^\infty \frac{1}{2^{3\cdot n}}$ .We have that

$S+1=\displaystyle\sum_{i=0}^\infty \frac{1}{2^{3\cdot n}}=\displaystyle\sum_{i=0}^\infty \frac{1}{{\left(2^3\right)}^n}=\frac{1}{1-\frac{1}{2^3}} \iff S=1-\frac{1}{1-\frac{1}{2^3}} \iff S=\frac{1}{7}$

As wanted.

It is very simple ; as we all know very popular formula i.e S = a/1 -r ( says equation 1 } according to the given question we have a = 1/8 amd b = 1/64 and r = b/a so; r= 1/8 hence put the value in the above formula so we get 1/8 / 1- 1/8 = so we get 1/7

Let S = \frac{1}{2^{3}}+\frac{1}{2^{6}}+\frac{1}{2^{9}}...

Therefore S = \frac{1}{8}+\frac{1}{8^{2}}+\frac{1}{8^{3}}...

We know that 8 = \frac{7}{8^{0}}+\frac{7}{8^{1}}+\frac{7}{8^{2}}+\frac{7}{8^{3}}...

So 8 = 7(\frac{1}{8^{0}}+\frac{1}{8^{1}}+\frac{1}{8^{2}}+\frac{1}{8^{3}}...)

8 = 7(1+S) \frac{8}{7} = 1 + S 1+\frac{1}{7} - 1 = S

S = \frac{1}{7}

$S=\frac{1}{2^3}+\frac{1}{2^6}+\dots \implies 2^3S=1+\frac{1}{2^3}+\frac{1}{2^6}+\dots \implies 8S=1+S \implies 7S=1 \implies S=\boxed{\large{\frac{1}{7}}}$

If you look at the pattern, each term is multiplied by (1/2^3) or 1/8. So r=1/8. Knowing the property of geometric series, we know this converges since |r|<1. Since the first term is given (1/8), we can use a/(1-r) to find the value of the series. So (1/8)/(7/8) is 1/7.

You can think about it in a logical way....as the denominator increases the value of the fraction tends more to zero so 1/7 is likely closer to zero then all the other numbers

$S=\frac {1}{2^3} + \frac {1}{2^6} + \frac {1}{2^9} + \cdots \newline 2^3 \times S=1+\frac{1}{2^3}+\frac{1}{2^6}+\cdots \newline 8S-S=7S=1 \Longrightarrow S=\frac{1}{7}$

Very simply put in this question you have to use the sum of the geometric series formula

Here we have $a_1= frac{1}{2^{3}}$ And $r=frac{1}{2^{3}}$ Put it in the formula and you'll get the answer

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Given Series, $\large \dfrac{1}{2^3} + \dfrac{1}{2^6} + \dfrac{1}{2^9} + \ldots$ It is an infinite Geometric progression, so, $\large S = \dfrac{a}{1 - r} = \dfrac{\frac{1}{2^3}}{1 - {\frac{1}{2^3}}} = \dfrac{1}{2^3 - 1} = \color{#3D99F6}{\boxed{\dfrac{1}{7}}}$

Let $S = \frac{1}{2^3} + \frac{1}{2^6} + \frac{1}{2^9} + \ldots$

$S = \frac{1}{2^3} ( \frac{1}{1} + \frac{1}{2^3} + \frac{1}{2^6} + \ldots )$

$S = \frac{1}{2^3} ( 1 + S )$

$S = \frac{1}{2^3} + \frac{1}{2^3} S$

$S = \frac{1}{8} + \frac{1}{8} S$

$8S = 1 + S$

$S = \frac{1}{7}$

This is an geometric series and it has an known formula

$S = \frac{1}{1-r}$ where

$r = \frac{a_{n+1}}{a_n}$

Our serie its

$\sum_{k=1}^{\infty} 2^{-3k}$

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

by exp propeties

$r=8^{-1}$

using the formula we have

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

solving it :

$S = 1/7$

The series is in G.P. Given that a=1÷8 ; r=1÷8

S∞= a÷(1-r) =1÷7

Using G.P. formula for sum to infinity. Where first term is 1/8 and common ratio is 1/8.

Meow. I love this question. Mammamiya.

Infinite GP series simply apply S=a÷(1-r) where a = 1st term of GP ,that is 1÷8 and r = common ratio , that is 1÷8

Using sum of GP of infinite terms a/1-r where a is first term and r is common difference