anonymous
  • anonymous
4+2{sum from 1 to infinity}4/(3^n) equals 4+2 * (4/3)/(1-(1/3)) What work was done between the first and second step?? D:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[4+2\sum_{1}^{\infty}4/3^n = 4+2(4/3)/(1-(1/3))\]
anonymous
  • anonymous
is this \[4+2\sum_{n=1}^{\infty}(\frac{4}{3})^n\] or \[4+2\sum_{k=1}^{\infty}\frac{4}{3^n}\]
anonymous
  • anonymous
Oops, sorry - still getting the hang of using the equation editor. It's the second one . . .

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anonymous
  • anonymous
ok then the 4 can come right out front and you have \[4+8\sum_{n=1}^{\infty}\frac{1}{3^n}\]
anonymous
  • anonymous
ignoring the 4 and the 8 you have \[\sum_{n=1}^{\infty}\frac{1}{3^n}\] and the formula for that is \[\frac{\frac{1}{3}}{1-\frac{1}{3}}\] is that part clear or no?
anonymous
  • anonymous
^ Why is that?
anonymous
  • anonymous
hmmm it take a bit of explaining. do you know how to sum a geometric series?
anonymous
  • anonymous
in general \[a+ar+ar^2+ar^3+...=\sum_{k=1}^{\infty}ar^k =\frac{a}{1-r}\] if \[0
anonymous
  • anonymous
Oh - never mind. I understand that part now :D
anonymous
  • anonymous
typo above, i should have written \[\sum_{k=0}^{\infty} ar^k=\frac{a}{1-r}\]
anonymous
  • anonymous
Thank you! I'm sorry I was a little slow. :)

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