Question

[Doraemon]

Tính 

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2004\times2005}+\frac{1}{2005\times2006}=A\)

\(\frac{1}{6}+\frac{2}{15}+\frac{4}{45}+\frac{2}{99}+\frac{10}{600}=A\)

Answer 1

[Ninh]

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)

\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)

Answer 2

[Phạm Tuấn Đạt]

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)

\(\Rightarrow A=1-\frac{1}{2006}\)

\(\Rightarrow A=\frac{2005}{2006}\)