Đáp án:
$C=\dfrac{1-\dfrac{1}{3^6}}{2}$
Giải thích các bước giải:
$C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^6}\\
\Rightarrow 3C=3\left ( \dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^6}\right )\\
\Leftrightarrow 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^5}\\
\Leftrightarrow 3C-C=\left (1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^5} \right )-\left ( \dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^6} \right )\\
\Leftrightarrow 2C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^5} -\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^6} \\
\Leftrightarrow 2C=1-\dfrac{1}{3^6} \\
\Leftrightarrow C=\dfrac{1-\dfrac{1}{3^6}}{2}$