Find the value of
$$\frac{1}{3+1}+\frac{2}{3^2+1}+\frac{4}{3^4+1}+\frac{8}{3^8+1}+\cdots+\frac{2^{2006}}{3^{2^{2006}}+1}$$
I try to form the summation into telescoping sum but I stuck.
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1https://math.stackexchange.com/q/1119245/42969 – Martin R Aug 30 '22 at 08:50
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As @MartinR shared, I think it is duplicated... – Vezen BU Aug 30 '22 at 08:51
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1Does this answer your 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 ... – Mark Bennet Aug 30 '22 at 09:20
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https://artofproblemsolving.com/community/c6h1759891p11501464 – anyon Aug 30 '22 at 09:38
1 Answers
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More generally, define $S_n = \frac{1}{3+1}+\frac{2}{3^2+1}+\frac{4}{3^4+1}+\frac{8}{3^8+1}+\cdots+\frac{2^{n}}{3^{2^{n}}+1}$.
Hint: $\frac{2^{n}}{3^{2^n}+1}=\frac{2^{n}}{3^{2^n}-1}-\frac{2^{n+1}}{3^{2^{n+1}}-1}$.
Another way is to use induction.
Vezen BU
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