bài tập tết của mik nè!

T

tranhainam1801

[TEX]A=(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})[/TEX]
[TEX]A=(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100})[/TEX]
[TEX]A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100})-(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})[/TEX]
[TEX]A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+ \frac{1}{100}[/TEX]
\Rightarrow [TEX]\frac{A}{B}=2015[/TEX]
 
Last edited by a moderator:
L

ledinhlocpt

A=$\frac{1}{b1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.............+\frac{1}{99}-\frac{1}{100}$
A=$(1+\frac{1}{3}+\frac{1}{5}+..........+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+.......\frac{1}{100})$
A=$(1+\frac{1}{3}+\frac{1}{5}+..........+\frac{1}{99})+(\frac{1}{2}+\frac{1}{4}+.......\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+.......\frac{1}{100})$
A=$(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50})$
A=$(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100})$
Mặt khác B=$\frac{2015}{51}+\frac{2015}{52}+\frac{2015}{53}+...+\frac{2015}{100}$
B=2015$(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100})$
=>$\frac{B}{A}$=2015
=>$\frac{B}{A}$ thuộc Z (đpcm)
 
Last edited by a moderator:
Top Bottom