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 Can someone show me how these two series are equal? im converting the series to a geometric series ar^n and I'm having a hard time with the algebra..

 

\(\sum_{n=1}^{infinity}\frac{{5}^{n+1}}{{3}^{n-1}} = \sum_{n=1}^{infinity}15(5/3)^n\)

 Nov 15, 2016
 #1
avatar+9673 
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\(\displaystyle\sum^{\infty}_{n=1}\dfrac{5^{n+1}}{3^{n-1}}=\displaystyle\sum^{\infty}_{n=1}\left(\dfrac{5^n}{3^n}\right)\left(\dfrac{5^1}{3^{-1}}\right)=\displaystyle\sum^{\infty}_{n=1}5\cdot3\left(\dfrac{5}{3}\right)^n=\displaystyle\sum^{\infty}_{n=1}15\left(\dfrac{5}{3}\right)^n\)

:) Hope this helps 

 Nov 15, 2016
 #2
avatar+270 
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It does, thanks! :3

gretzu  Nov 15, 2016

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