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zmudz
 one year ago
A closed form for the sum \( S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} \)
is \(1  \frac{a^{n+b}}{3^{2^{n+c}}1},\) where \(a, b,\) and \(c\) are integers. Find \(a+b+c\).
zmudz
 one year ago
A closed form for the sum \( S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} \) is \(1  \frac{a^{n+b}}{3^{2^{n+c}}1},\) where \(a, b,\) and \(c\) are integers. Find \(a+b+c\).

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$S_n=\sum_{k=0}^n\frac{2^{k+1}}{3^{2^k}+1}$$consider $$S_0=\frac12\\S_1=\frac12+\frac25=\frac9{10}\\S_2=\frac12+\frac25+\frac4{41}=\frac{409}{410}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0according to their formula we have $$S_0=1\frac{a^b}{3^{2^c}1}\\\frac12=\frac{a^b}{3^{2^c}1}\\3^{2^c}1=2a^b$$ similarly $$S_1=1\frac{a^{1+b}}{3^{2^{1+c}}1}\\\frac1{10}=\frac{a^{1+b}}{3^{2^{1+c}}1}\\3^{2^{1+c}}1=10a^{1+b}$$and $$S_2=1\frac{a^{2+b}}{3^{2^{2+c}}1}\\\frac1{410}=\frac{a^{2+b}}{3^{2^{2+c}}1}\\3^{2^{2+c}}1=410a^{2+b}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we get $$5a=\frac{(3^{2^c})^21}{3^{2^c}1}=3^{2^c}+1\\41a=\frac{(3^{2^{1+c}})^21}{3^{2^{1+c}}1}=3^{2^{1+c}}+1$$so $$36a=3^{2^{1+c}}3^{2^c}=3^{2^c}(3^{2^c}1)=3^{2^c}\cdot 2a^b$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider \(36=6^2=2^2\cdot 3^2\) so $$2\cdot 3^2 a=3^{2^c}\cdot a^b$$ implying that a solution exists where \(c=1\), so $$2a=a^b$$ so take \(b=2\), \(a=2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thus \(a+b+c=2+2+1=5\)
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