Question

Tìm x là số tự nhiên sao cho: \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1000}{2002}\)

Answer 1

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1000}{2002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1000}{2002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2002}\)

<=>x+1=2002

=>x=2001

Answer 2

dap an :

X=X