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$$\sum_{k=1}^\infty \frac{2\times 3^k}{4^{2k+1}}$$

Hi all, I finally am getting the hang of MathJax (sort of) thank goodness! I was hoping for some help on a problem involving series. I am stuck trying to re-write this problem to make it easier to solve. For example, I was hoping to use the fact that a Sum = $$S\infty= \frac{a_1}{1-r}$$
to solve but unlike a problem with simply k+1 in the denominator, this has a constant infront of it which is throwing me off.. any tips on how approach solving or a trick that I am unaware of?

Thank you!

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    $\begingroup$ Next stop: getting the hang of titles! :) The key directive is that people should be able to know something about your question without having to open it. $\endgroup$ – Asaf Karagila Aug 5 '18 at 7:53
  • $\begingroup$ Thank you Asaf, had no idea we could directly integrate problems in the title and have mathjax pick it up! $\endgroup$ – jackbenimbo Aug 5 '18 at 20:51
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This is a geometric series $$\sum_{k=1}^\infty \frac{2*3^k}{4^{2k+1}}=\sum_{k=1}^\infty \dfrac12\left(\frac{3}{16}\right)^{k}$$ with $a_1=\dfrac12\dfrac{3}{16}$ and $q=\dfrac{3}{16}$, then $$S_\infty=\dfrac{\dfrac12\dfrac{3}{16}}{1-\dfrac{3}{16}}=\dfrac{3}{26}$$

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  • $\begingroup$ so you reduced 2/4, how are you getting 3/16 though? (Thanks in advance) $\endgroup$ – jackbenimbo Aug 5 '18 at 6:31
  • $\begingroup$ the power $4$ is $2k+1$, $(4^2)^k*4$. $\endgroup$ – Nosrati Aug 5 '18 at 6:32
  • $\begingroup$ I see it clearly now thank you so much for your time on a Saturday evening! Cheers $\endgroup$ – jackbenimbo Aug 5 '18 at 6:41
  • $\begingroup$ you are welcome. $\endgroup$ – Nosrati Aug 5 '18 at 6:41

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