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Calculate the value of $\mathrm{n}$ if: $\sum_{k=1}^{n} 3^{k}=1092$
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This is a geometric sum, and we know the sum will be given by

$\sum_{k=1}^n 3^k = \frac{a(r^n-1)}{r-1}$

Plugging k=1 into $3^k$ we get a = 3, and we see that r=3

Hence we have, $\sum_{k=1}^n 3^k = \frac{3(3^n-1)}{3-1} = \frac{3(3^n-1)}{2} ...(1)$

but we are given that $\sum_{k=1}^n 3^k = 1092 ...(2)$

So we can equate (1) and (2) to solve for n

$\frac{3(3^n-1)}{2} =1092$

$3(3^n-1) = 2184$

$3^n-1= 728$

$3^n=729$

$3^n = 3^6$

$\implies n=6$
by Wooden (1,332 points)

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