Question

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 \(a+b+c\).

Answer 1

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Answer 2

$$S_n=\sum_{k=0}^n\frac{2^{k+1}}{3^{2^k}+1}$$consider $$S_0=\frac12\\S_1=\frac12+\frac25=\frac9{10}\\S_2=\frac12+\frac25+\frac4{41}=\frac{409}{410}$$

Answer 3

according 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 $$S_2=1-\frac{a^{2+b}}{3^{2^{2+c}}-1}\\\frac1{410}=\frac{a^{2+b}}{3^{2^{2+c}}-1}\\3^{2^{2+c}}-1=410a^{2+b}$$

Answer 4

so we get $$5a=\frac{(3^{2^c})^2-1}{3^{2^c}-1}=3^{2^c}+1\\41a=\frac{(3^{2^{1+c}})^2-1}{3^{2^{1+c}}-1}=3^{2^{1+c}}+1$$so $$36a=3^{2^{1+c}}-3^{2^c}=3^{2^c}(3^{2^c}-1)=3^{2^c}\cdot 2a^b$$

Answer 5

consider \(36=6^2=2^2\cdot 3^2\) 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 \(b=2\), \(a=2\)

Answer 6

thus \(a+b+c=2+2+1=5\)