Question

[Nguyễn Lu]

\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{308}+\frac{1}{309}\)

\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+....+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)

Tính\(\frac{\:A}{B}\)

Answer 1

[Thanh Tùng DZ]

Ta có :

\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)

\(B=\left(\frac{307}{2}+1\right)+\left(\frac{306}{3}+1\right)+...+\left(\frac{3}{306}+1\right)+\left(\frac{2}{307}+1\right)+\left(\frac{1}{308}+1\right)+1\)

\(B=\frac{309}{2}+\frac{309}{3}+...+\frac{309}{306}+\frac{309}{307}+\frac{309}{308}+\frac{309}{309}\)

\(B=309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{306}+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}}{309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{308}+\frac{1}{309}\right)}\)

\(\frac{A}{B}=\frac{1}{309}\)