# How do you find the sum of the infinite geometric series 12 + 6 + 3 + . . . ?

Nov 9, 2015

24

#### Explanation:

Let the series be

${S}_{1} = {\sum}_{i = 1 \to \infty} {a}_{i} = 12 + 6 + 3 + \ldots . + {a}_{n} + \ldots + {a}_{\infty}$

Consider the generic case: of ${S}_{2}$

Let k be any constant
Let r be the geometric ratio

Then ${a}_{i} = k {r}^{i}$

color(blue)("Then "S_2 = kr^1 + kr^2 + kr^3 + ... + kr^n......... ( 1 )

Note that you could have $\text{ } k {r}^{0} + k {r}^{1} + k {r}^{2} + \ldots + k {r}^{n}$

Multiply (1) by r giving:

color(blue)( rS_2= kr^2 + kr^3 + kr^4 + ... +kr^n + kr^(n+1)......... (2)

(2) - (1) ginves:

$\textcolor{b r o w n}{r {S}_{2} - {S}_{2} = - k {r}^{1} + k {r}^{n + 1}}$

color(brown)(S_2 =( kr(r^n -1))/(r-1)........( 3 )

Notice that the number sequence is reducing implying the $r < 1$

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To find r consider ${a}_{1} / {a}_{2} = \frac{k r}{k {r}^{2}} = \frac{1}{r} = \frac{12}{6}$

Inverting $\frac{1}{r}$ gives $\textcolor{red}{r = \frac{6}{12} = \frac{1}{2}}$
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To find k
$k r - k {r}^{2} = k \left(r - {r}^{2}\right) = k \left(\frac{1}{2} - \frac{1}{4}\right) = \frac{1}{4} k = 12 - 6$

$\frac{1}{4} k = 6$

so $\textcolor{red}{k = 24}$
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$\textcolor{red}{\text{Check}}$
${S}_{2} = k {r}^{1} + k {r}^{2} + k {r}^{3} \to \left(24 \times \frac{1}{2}\right) + \left(24 \times \frac{1}{4}\right) + \left(24 \times \frac{1}{8}\right) \ldots . .$

${S}_{2} = 12 + 6 + 3 + . .$
this matches so it is safe to assume we have found the system.
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Plug values for k and r in equation (3) and determine limits

${\lim}_{n \to \infty} \frac{24 \left({\left(\frac{1}{2}\right)}^{n + 1} - r\right)}{\frac{1}{2} - 1}$

As $n \to \infty , {\left(\frac{1}{2}\right)}^{n + 1} \to 0$ so we end up with $- 24 \frac{r}{- \left(\frac{1}{2}\right)} = - 24 \frac{\frac{1}{2}}{- \frac{1}{2}}$

$\textcolor{red}{{S}_{1} = + 24}$