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\(C=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\)
\(C=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(C=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(D=\frac{1}{51.100}+\frac{1}{52.99}+\frac{1}{53.98}+...+\frac{1}{99.52}+\frac{1}{100.51}\)
\(D=\frac{1}{151}.\left(\frac{151}{51.100}+\frac{151}{52.99}+\frac{151}{53.98}+...+\frac{151}{99.52}+\frac{151}{100.51}\right)\)
\(D=\frac{1}{151}.\left(\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+...+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\right)\)
\(D=\frac{1}{151}.\left(\frac{2}{100}+\frac{2}{99}+...+\frac{2}{51}\right)\)
\(D=\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)\)
\(\Rightarrow C:D=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)}\)
\(\Rightarrow C:D=\frac{151}{2}=75\frac{1}{2}\)
b: \(F=2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{240}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(=2\cdot\dfrac{7}{16}=\dfrac{7}{8}\)
c: \(3G=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\)
\(\Leftrightarrow2G=1-\dfrac{1}{243}=\dfrac{242}{243}\)
hay G=121/243
d: \(2H=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Leftrightarrow H=1-\dfrac{1}{1024}\)
2F= 1-1/2+1/2^2+...+1/2^98-1/2^99
2F+F= 1-1/2^100
3F=2^100-1 / 2^100
F=2^100-1 / 2^100.3
Vậy F=2^100-1 / 2^100.3
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