Cho

Câu hỏi: Cho $A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{200}$ ; $B = \frac{1}{199} + \frac{2}{198} + \frac{3}{197} + ... + \frac{198}{2} + \frac{199}{1}$ . Tính $\frac{A}{B}$

Đáp án

- Hướng dẫn giải

$\begin{array}{l} B = \frac{1}{199} + \frac{2}{198} + \frac{3}{197} + ... + \frac{198}{2} + \frac{199}{1} \\ = (\frac{1}{199} + 1) + (\frac{2}{198} + 1) + ... + (\frac{198}{2} + 1) + 1 \\ = \frac{200}{199} + \frac{200}{198} + ... + \frac{200}{2} + \frac{200}{200} \\ = 200 (\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{200}) \\ = 200 A \\ \Rightarrow \frac{B}{A} = \frac{200 A}{A} = 200 \end{array}$