Câu hỏi của Nguyễn Khánh Ngoc

Question

Cho A=457/1+456/2+455/3+…+1/457, Chứng minh: A>2016

ミ★Zero ❄ ( Hoàng Nhật ) · Accepted Answer

$A=\left(\dfrac{456}{2}+1\right)+...+\left(\dfrac{2}{456}+1\right)+\left(\dfrac{1}{457}+1\right)+1$

$A=458+\dfrac{458}{2}+....+\dfrac{458}{456}+\dfrac{458}{457}-\dfrac{458}{458}$

$A=458\left(\dfrac{1}{2}+...+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\right)$

Ta xét $\dfrac{1}{2}+....+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}$có :

$\dfrac{1}{2}=\dfrac{1}{2}$

$\dfrac{1}{3}+\dfrac{1}{4}>\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}$

$\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{8}>\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}=\dfrac{1}{2}$

$\dfrac{1}{9}+\dfrac{1}{10}+....+\dfrac{1}{16}>\dfrac{1}{16}+....+\dfrac{1}{16}=\dfrac{1}{2}$

$\dfrac{1}{17}+\dfrac{1}{18}+....+\dfrac{1}{32}>\dfrac{1}{32}+.....+\dfrac{1}{32}=\dfrac{1}{2}$

$\dfrac{1}{33}+\dfrac{1}{34}+....+\dfrac{1}{64}>\dfrac{1}{64}+....+\dfrac{1}{64}=\dfrac{1}{2}$

$\dfrac{1}{65}+\dfrac{1}{66}+.....+\dfrac{1}{128}>\dfrac{1}{128}+....+\dfrac{1}{128}=\dfrac{1}{2}$

$\dfrac{1}{129}+\dfrac{1}{130}+.....+\dfrac{1}{256}>\dfrac{1}{256}+....+\dfrac{1}{256}=\dfrac{1}{2}$

$\dfrac{1}{257}+\dfrac{1}{258}+....+\dfrac{1}{458}>\dfrac{1}{458}+...+\dfrac{1}{458}=\dfrac{1}{2}$

Vậy ta thấy được rằng

$\dfrac{1}{2}+...+\dfrac{1}{456}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{202}{458}$

$=4+\dfrac{202}{458}=\dfrac{2034}{458}$

Vậy $A>458.\dfrac{2034}{458}=2034$

Hay tức là A > 2016 ( đpcm )

Câu trả lời: $A=\left(\dfrac{456}{2}+1\right)+...+\left(\dfrac{2}{456}+1\right)+\left(\dfrac{1}{457}+1\right)+1$