已知 $a_k=\dfrac{2^k}{3^{2^k}+1}$，$k\in \mathbb{N}^{*}$，$S_n=a_1+a_2+\cdots+a_n$，$T_n=a_1a_2\cdots a_n$，求 $\dfrac{S_9}{T_9}$ 的值．
【难度】
【出处】
无
【标注】
【答案】
$\dfrac{3^{1024}-4097}{2^{50}}$
【解析】
因为\[\begin{split}
a_k
&=\dfrac{2^k}{3^{2^k}+1}\\
&=\dfrac{2^k\left(3^{2^k}-1\right)}{\left(3^{2^k}+1\right)\left(3^{2^k}-1\right)}\\
&=\dfrac{2^k\left(3^{2^k}+1\right)-2^{k+1}}{\left(3^{2^k}+1\right)\left(3^{2^k}-1\right)}\\
&=\dfrac{2^k}{3^{2^k}-1}-\dfrac{2^{k+1}}{3^{2^{k+1}}-1},
\end{split}\]所以\[S_n=\dfrac{1}{4}-\dfrac{2^{n+1}}{3^{2^{n+1}}-1}.\]又因为\[\begin{split}T_n
&=a_1a_2\cdots a_n\\
&=\dfrac{2^{\frac{n(n+1)}{2}}}{\left(3^{2^1}+1\right)\left(3^{2^2}+1\right)\cdots\left(3^{2^n}+1\right)}\\
&=\dfrac{\left(3^{2^1}-1\right)\cdot 2^{\frac{n(n+1)}{2}}}{\left(3^{2^1}-1\right)\left(3^{2^1}+1\right)\left(3^{2^2}+1\right)\cdots\left(3^{2^n}+1\right)}\\
&=\dfrac{2^{\frac{n^2+n+6}{2}}}{3^{2^{n+1}}-1},
\end{split}\]所以\[
\dfrac{S_9}{T_9}=\dfrac{\dfrac{3^{1024}-1}{4}-1024}{2^{48}}=\dfrac{3^{1024}-4097}{2^{50}}.\]
a_k
&=\dfrac{2^k}{3^{2^k}+1}\\
&=\dfrac{2^k\left(3^{2^k}-1\right)}{\left(3^{2^k}+1\right)\left(3^{2^k}-1\right)}\\
&=\dfrac{2^k\left(3^{2^k}+1\right)-2^{k+1}}{\left(3^{2^k}+1\right)\left(3^{2^k}-1\right)}\\
&=\dfrac{2^k}{3^{2^k}-1}-\dfrac{2^{k+1}}{3^{2^{k+1}}-1},
\end{split}\]所以\[S_n=\dfrac{1}{4}-\dfrac{2^{n+1}}{3^{2^{n+1}}-1}.\]又因为\[\begin{split}T_n
&=a_1a_2\cdots a_n\\
&=\dfrac{2^{\frac{n(n+1)}{2}}}{\left(3^{2^1}+1\right)\left(3^{2^2}+1\right)\cdots\left(3^{2^n}+1\right)}\\
&=\dfrac{\left(3^{2^1}-1\right)\cdot 2^{\frac{n(n+1)}{2}}}{\left(3^{2^1}-1\right)\left(3^{2^1}+1\right)\left(3^{2^2}+1\right)\cdots\left(3^{2^n}+1\right)}\\
&=\dfrac{2^{\frac{n^2+n+6}{2}}}{3^{2^{n+1}}-1},
\end{split}\]所以\[
\dfrac{S_9}{T_9}=\dfrac{\dfrac{3^{1024}-1}{4}-1024}{2^{48}}=\dfrac{3^{1024}-4097}{2^{50}}.\]
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