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anonymous
 5 years ago
Find the sum of the infinite geometric series. 1/3^7 + 1/3^9 + 1/3^11 + 1/3^13...
anonymous
 5 years ago
Find the sum of the infinite geometric series. 1/3^7 + 1/3^9 + 1/3^11 + 1/3^13...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The form of the nth term is\[a_n=\frac{1}{9^{n}}\]for the series\[\frac{1}{3^7}\sum_{n=0}^{\infty}\frac{1}{3^{2n}}=\frac{1}{3^7}\sum_{n=0}^{\infty}\frac{1}{9^{n}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let the partial sum of the first a_n terms be\[s_n=1+\frac{1}{9}+\frac{1}{9^2}+...+\frac{1}{9^{n1}}\]Then\[\frac{1}{9}s_n=\frac{1}{9}+\frac{1}{9^2}+\frac{1}{9^3}+...+\frac{1}{9^n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Subtract the second from the first:\[(1\frac{1}{9})s_n=1\frac{1}{9^n}\rightarrow s_n=\frac{1\frac{1}{9^n}}{8/9}=\frac{9}{8}(1\frac{1}{9^n})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The series will be the limit of the following sequence of partial sums:\[\lim_{n \rightarrow \infty}\frac{1}{3^7}s_n=\lim_{n \rightarrow \infty}\frac{1}{3^7}.\frac{9}{8}.(1\frac{1}{9^n})=\frac{9}{8.3^7}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3^7}+\frac{1}{3^9}+...=\frac{9}{8.3^7}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whats that fraction simplified?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{9}{8.3^7}=\frac{1}{1944}\]
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