Question

[yoring]

Tìm n biết :\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+......+\(\frac{2}{n.\left(n+1\right)}\)=\(\frac{2003}{2004}\)

Answer 1

[Lightning Farron]

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2003}{2004}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)\(n+1=4008\Rightarrow n=4007\)

Answer 2

[yoring]

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