Question

 C=(1/2+1/3+ 1/4+... +1/100): (99/1 + 98/2 + 97/3 ... +1/99)

tìm C

 

Answer 1

Ta có: \(\frac{99}{1}+\frac{98}{2}+\cdots+\frac{1}{99}\)

\(=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+\cdots+\left(1+\frac{1}{99}\right)+1\)

\(=\frac{100}{2}+\frac{100}{3}+\cdots+\frac{100}{99}+\frac{100}{100}\)

\(=100\left(\frac12+\frac13+\cdots+\frac{1}{100}\right)\)

Ta có: \(C=\frac{\frac12+\frac13+\frac14+\cdots+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\cdots+\frac{1}{99}}\)

\(=\frac{\frac12+\frac13+\cdots+\frac{1}{100}}{100\left(\frac12+\frac13+\cdots+\frac{1}{100}\right)}=\frac{1}{100}\)