Question

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+....+\frac{1}{204}}\)

Answer 1

\(\)Đặt \(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+...+\frac{1}{204}}=\frac{B}{C}\)

Biến đổi C:

\(C=\left(\frac{204}{1}+1\right)+\left(\frac{203}{2}+1\right)+\left(\frac{202}{3}+1\right)+...+\left(\frac{1}{204}+1\right)-204\)

\(=205+\frac{205}{2}+\frac{205}{3}+..+\frac{205}{204}+\frac{205}{205}-205\)

\(=205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}}{205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)}=\frac{1}{205}\)