Đáp án:
Giải thích các bước giải:
Đặt \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2001}-\dfrac{1}{2002}\\ Q=\dfrac{1}{1002}+\dfrac{1}{1003}...+\dfrac{1}{2002}\)
Ta có:
\(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2001}-\dfrac{1}{2002}\\ \Rightarrow P=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{2001}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2002}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2002}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2002}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2002}\right)\)\(\Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2002}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2002}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2002}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{1001}\right)\\ \Rightarrow P=\dfrac{1}{1002}+\dfrac{1}{1003}+...+\dfrac{1}{2002}\\ \Rightarrow P=Q\)\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2001}-\dfrac{1}{2002}=\dfrac{1}{1002}+\dfrac{1}{1003}...+\dfrac{1}{2002}\left(đpcm\right)\)
