Question

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+....+\dfrac{1}{99}+\dfrac{1}{x.\left(x+1\right)}=\dfrac{1}{2001}\)

Answer 1

Giải:

Ta có:

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{99}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{2001}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{2001}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{2001}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1}{2001}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1999}{4002}\)

\(\Leftrightarrow x+1=\dfrac{4002}{1999}\)

\(\Leftrightarrow x=\dfrac{2003}{1999}\)

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