\(\frac{2}{3x5}+\frac{2}{5.7}+\frac{2}{7x9}+...+\frac{2}{99x101}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
\(\frac{\Rightarrow1}{15}+\frac{1}{35}+...+\frac{1}{9999}=\frac{98}{303x2}=\frac{49}{303}\)
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