How do you find the sum of the first 5 terms of the geometric series: 4+ 16 + 64…?

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How do you find the sum of the first 5 terms of the geometric series: 4+ 16 + 64…?

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Answer 1 · 2015-12-29T18:00:28

$1364$ $1364$

Explanation:

$\frac{a_{2}}{a_{1}} = \frac{16}{4} = 4$ $a_2/a_1=16/4=4$
$\frac{a_{3}}{a_{2}} = \frac{64}{16} = 4$ $a_3/a_2=64/16=4$

$\Rightarrow$ $implies$ common ratio $= r = 4$ $=r=4$ and $a_{1} = 4$ $a_1=4$

Sum of a geometric series is given by
$S u m = S = \frac{a (1 - r^{n})}{1 - r}$ $Sum=S=(a(1-r^n))/(1-r)$

Where $a$ $a$ is the first term $r$ $r$ is the common ratio and $n$ $n$ is the number of terms.

$S u m = S = \frac{4 (1 - 4^{5})}{1 - 4} = \frac{4 (1 - 1024)}{- 3} = \frac{4 (- 1023)}{- 3} = \frac{- 4092}{- 3} = 1364$ $Sum=S=(4(1-4^5))/(1-4)=(4(1-1024))/(-3)=(4(-1023))/(-3)=(-4092)/(-3)=1364$

Hence the required sum is $1364$ $1364$ .

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How do you find the sum of the first 5 terms of the geometric series: 4+ 16 + 64…?

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