1/1 . 2 + 1/ 3 . 4 + 1/5 . 6 + ...+ 1/99 . 100
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/99 - 1/100
= ( 1 + 1/3 + 1/5 + ...+ 1/99 ) - ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - 2 . ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - ( 1 + 1/2 + ...+ 1/50 )
= 1/51 + 1/52 + ...+ 1/100
Tham khảo nha !!!
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\) (đpcm)
