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I'm stuck on the question $\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^{k+3}$

I know that $\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k$ is solved by using $\sum_{k=0}^{\infty} a^{k} =$ $\frac{1}{1-a}$ and the answer is $\frac{3}{2}$

So is there a way I could apply that to the above question or is there a different way to approach the problem?

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    $\begingroup$ Hint: $$\left(\frac{1}{3}\right)^{k+3}=\frac{1}{27}\left(\frac{1}{3}\right)^k$$ $\endgroup$ – Ángel Mario Gallegos Jan 25 '16 at 4:36
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    $\begingroup$ Write the sum as $\sum_{k=0}^{\infty} \frac{1}{3}^k \ \cdot \ \left( \frac{1}{3} \right)^3 \ = \ \frac{1}{27} \ \sum_{k=0}^{\infty} \frac{1}{3}^k \ $ by applying rules of exponents. $\endgroup$ – colormegone Jan 25 '16 at 4:37
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$$\sum_{k=0}^n\left(\frac{1}{3}\right)^{k+3}=\frac{1}{3^3}\sum_{k=0}^n\left(\frac{1}{3}\right)^{k}\to\frac{1}{27}\frac{1}{1-\frac{1}{3}}=\frac{1}{27}\frac{3}{2}=\frac{1}{18}$$

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  • $\begingroup$ Could I also apply that to the sum with $\frac{1}{3}^{2k+1}$? How would I take care of that 2? $\endgroup$ – Lindsey G Jan 25 '16 at 4:47
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    $\begingroup$ $(\frac{1}{3})^{2k+1}=\frac{1}{3}(\frac{1}{3})^{2k}=\frac{1}{3} \left[\left(\frac{1}{3}\right)^2\right]^k=\frac{1}{3}(\frac{1}{9})^k$ $\endgroup$ – sinbadh Jan 25 '16 at 4:49

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