回復 17# 阿光 的帖子
第14題
\(\begin{align}
& \sum\limits_{n=1}^{\infty }{\left\{ \frac{\sum\limits_{k=1}^{{{2}^{n}}-1}{\left[ {{\log }_{2}}k \right]}}{{{3}^{n}}} \right\}} \\
& =\frac{1\times 2}{3{}^{2}}+\frac{1\times 2+2\times 4}{3{}^{3}}+\frac{1\times 2+2\times 4+3\times 8}{3{}^{4}}+...+\frac{1\times 2+2\times 4+3\times 8+...+\left( n-1 \right)\times {{2}^{n-1}}}{3{}^{n}} \\
& =\sum\limits_{n=2}^{\infty }{\frac{\left( n-2 \right)\times {{2}^{n}}+2}{{{3}^{n}}}} \\
& =\sum\limits_{n=2}^{\infty }{\left( n-2 \right)\times {{\left( \frac{2}{3} \right)}^{n}}}+2\sum\limits_{n=2}^{\infty }{\frac{1}{{{3}^{n}}}} \\
& =\frac{8}{3}+\frac{1}{3} \\
& =3 \\
\end{align}\)