Question

(1/51+1/52+1/53+....+1/100) : (1/1x2+1/3x4+...+1/99x100)

Answer 1

Có \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}=+....+\frac{1}{99}-\frac{1}{100}\)

\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

= \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=1\)