planhetkk

2022-09-01

Find sum of geometric series 2+6+18+...+1458

Nolan Tyler

Beginner2022-09-02Added 9 answers

Given that first term ${a}_{1}=2$

Common ratio $q=\frac{{a}_{2}}{{a}_{1}}=\frac{6}{2}=3$

General expression for nth term is

$a}_{n}={a}_{1}\cdot {q}^{n-1$

Put the above equation's last term's value in to find its number

$1458=2\cdot {3}^{n-1}$

$\Rightarrow {3}^{n-1}=\frac{1458}{2}=729$

When we use 729 as a power of 3, we get

$3}^{n-1}={3}^{6$, comparing the exponents

n−1=6

or n=7

Now, the expression gives the sum of n terms.

$S}_{n}=\frac{{a}_{1}(1-{q}^{n})}{1-q$

$S}_{7}=\frac{2(1-{3}^{7})}{1-3$

$=\frac{2(1-2187)}{-2}=2186$

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