How do you determine the sum of 2 + 8 + 32 + ... + 131 072?

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How do you determine the sum of 2 + 8 + 32 + ... + 131 072?

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Answer 1 · 2015-12-30T08:08:20

$S u m = 174762$ $Sum=174762$

Explanation:

$\frac{a_{2}}{a_{1}} = \frac{8}{2} = 4$ $a_2/a_1=8/2=4$
$\frac{a_{3}}{a_{2}} = \frac{32}{8} = 4$ $a_3/a_2=32/8=4$
$\Rightarrow$ $implies$ common ration $= r = 4$ $=r=4$ and the series is geometric.

Nth term of a geometric series is given by

$T_{n} = a_{1} r^{n - 1}$ $T_n=a_1r^(n-1)$

Where $T_{n}$ $T_n$ is the nth term, $a_{1}$ $a_1$ is the first term, $r$ $r$ is the common ration and $n$ $n$ is the number of terms.

Here $a_{1} = 2$ $a_1=2$ , $r = 4$ $r=4$ Last term $= 131072$ $=131072$

Let last term be the nth term

$\Rightarrow 131072 = 2 \cdot {(4)}^{n - 1}$ $implies 131072=2*(4)^(n-1)$
$\Rightarrow 4^{n - 1} = 65536$ $implies 4^(n-1)=65536$
$\Rightarrow 4^{n - 1} = 4^{8}$ $implies 4^(n-1)=4^8$
$\Rightarrow (n - 1) = 8$ $implies (n-1)=8$
$\Rightarrow n = 9$ $implies n=9$

Sum of geometric series is given by
$S u m = \frac{a_{1} (1 - r^{n})}{1 - r}$ $Sum=(a_1(1-r^n))/(1-r)$
Where $r$ $r$ is the common ratio $a_{1}$ $a_1$ is the first term and $n$ $n$ is the number of terms.

$\Rightarrow S u m = \frac{2 (1 - 4^{9})}{1 - 4} = \frac{2 (1 - 262144)}{- 3} = \frac{2 (- 262143)}{- 3} = \frac{- 524286}{- 3} = 174762$ $implies Sum=(2(1-4^9))/(1-4)=(2(1-262144))/-3=(2(-262143))/-3=(-524286)/-3=174762$

$\Rightarrow S u m = 174762$ $implies Sum=174762$ .

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How do you determine the sum of 2 + 8 + 32 + ... + 131 072?

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