Question

[Hạnh Nguyễn thị]

$A=$

$\genfrac{}{}{0ex}{}{1}{1.2}+\genfrac{}{}{0ex}{}{1}{3.4}+\genfrac{}{}{0ex}{}{1}{5.6}+\dots +\genfrac{}{}{0ex}{}{1}{49.50}$. Chứng minh rằng $A<1$.

chứng minh là j

Answer 1

[Akai Haruma]

$A=1.21+3.41+...+49.501$ hiển nhiên $>1$ rồi mà bạn. Bạn xem lại đề.

Answer 2

[Nguyễn Thị Thương Hoài]

A = ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{5.6}\)} + ... + ^{\(\dfrac{1}{49.50}\)} 

A <  ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{5.6}\)} + ... + ^{\(\dfrac{1}{49.50}\)} +^{\(\dfrac{1}{2.3}+\dfrac{1}{4.5}+\dfrac{1}{6.7}\)+...+\(\dfrac{1}{48.49}\)}

A < ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{2.3}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{4.5}\)}+^{\(\dfrac{1}{5.6}\)} +^{\(\dfrac{1}{6.7}\)}+.. +^{\(\dfrac{1}{47.48}\)}+ ^{\(\dfrac{1}{48.49}\)}+ ^{\(\dfrac{1}{49.50}\)} 

A < ^{\(\dfrac{1}{1}\)}-^{\(\dfrac{1}{2}\)} + ^{\(\dfrac{1}{2}\)} - ^{\(\dfrac{1}{3}\)} + ^{\(\dfrac{1}{3}\)} - ^{\(\dfrac{1}{4}\)} + ... + ^{\(\dfrac{1}{49}\)} - ^{\(\dfrac{1}{50}\)}

A < ^{\(\dfrac{1}{1}\)} - ^{\(\dfrac{1}{50}\)} < 1 (đpcm)