How do you find sum of geometric series 2+6+18+...+1458?

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How do you find sum of geometric series 2+6+18+...+1458?

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Answer 1 · 2016-03-15T08:02:03

$2186$ $2186$

Explanation:

Given that first term $a_{1} = 2$ $a_1=2$

Common ratio $q = \frac{a_{2}}{a_{1}} = \frac{6}{2} = 3$ $q=a_2/a_1=6/2=3$

General expression for $n t h$ $nth$ term is

$a_{n} = a_{1} \cdot q^{n - 1}$ $a_n=a_1*q^(n-1)$

To ascertain what is number of the last term, insert in the above equation

$1458 = 2 \cdot 3^{n - 1}$ $1458=2*3^(n-1)$
$\Rightarrow 3^{n - 1} = \frac{1458}{2} = 729$ $=> 3^(n-1) = 1458/2=729$
Writing $729$ $729$ as power of $3$ $3$ we obtain

$3^{n - 1} = 3^{6}$ $3^(n-1) = 3^6$ , comparing the exponents
$n - 1 = 6$ $n-1=6$
or $n = 7$ $n=7$
Now sum of $n$ $n$ terms is given by the expression

$S_{n} = \frac{a_{1} (1 - q^{n})}{1 - q}$ $S_n = (a_1(1-q^n))/(1-q)$

$S_{7} = \frac{2 (1 - 3^{7})}{1 - 3}$ $S_7 = (2 (1-3^7)) / (1-3)$
$= \frac{2 (1 - 2187)}{- 2} = 2186$ $= (2 (1-2187))/-2 = 2186$

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How do you find sum of geometric series 2+6+18+...+1458?

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