Giải:
a) \(F=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{240}\)
\(\Leftrightarrow F=\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{15.16}\)
\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{15}-\dfrac{1}{16}\)
\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{16}\)
\(\Leftrightarrow F=\dfrac{7}{16}\)
Vậy ...
b) \(G=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
\(\Leftrightarrow G=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\)
\(\Leftrightarrow\dfrac{1}{3}G=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}+\dfrac{1}{3^6}\)
\(\Leftrightarrow\dfrac{2}{3}G=\dfrac{1}{3}-\dfrac{1}{3^6}\)
\(\Leftrightarrow G=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^6}}{\dfrac{2}{3}}\)
\(\Leftrightarrow G=\dfrac{\left(\dfrac{1}{3}-\dfrac{1}{3^6}\right)3}{2}\)
\(\Leftrightarrow G=\dfrac{1-\dfrac{1}{3^5}}{2}\)
\(\Leftrightarrow G=\dfrac{\dfrac{3^5-1}{3^5}}{2}\)
\(\Leftrightarrow G=\dfrac{3^5-1}{3^5.2}\)
Vậy ...
c) Tương tự b)
