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

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!

  • 1
    $\begingroup$ 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

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 $$S_\infty=\dfrac{\dfrac12\dfrac{3}{16}}{1-\dfrac{3}{16}}=\dfrac{3}{26}$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.