Question

[Trần Duy Tùng]

Cho \(A=\dfrac{1}{1.21}+\dfrac{1}{2.22}+\dfrac{1}{3.23}+...+\dfrac{1}{80.100}\);

\(B=\dfrac{1}{1.81}+\dfrac{1}{2.82}+\dfrac{1}{3.83}+...+\dfrac{1}{20.100}\).

Tính \(\dfrac{A}{B}\).

Answer 1

[Nguyễn An Ninh]

A=20/1.21+20/2.22+...+20/80.100

=1-1/21+1/2-1/22+...+1/80-1/100

=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)

80B=80/1.81+80/2.82+...+8/20.100

=1-1/81+1/2-1/82+...+1/20-1/100

=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)

=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)

=>20A=80B

=>A=4B