chứng minh dãy phân số có quy luật

N

ngocsangnam12

Chứng minh rằng:
$B= \frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-..........-\frac{1}{7^{100}}<\frac{1}{50}$
Ta có:
$7^2 B$= $\frac{1}{7}-(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-.............-\frac{1}{7^{96}}+\frac{1}{7^{98}})$
+
$B$= $(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-..........+\frac{1}{7^{98}})-\frac{1}{7^{100}}$
_____________________________________________________

$=> 50B$= $\frac{1}{7}-\frac{1}{7^{100}}$
$=> B=$ =$\frac{7^{100}-7}{7^{101}.50}=\frac{7.(7^{99-1})}{7.7^{100}.50}=\frac{7^{99}-1}{7^{100}.50}<\frac{7^{100}}{7^{100}.50}=\frac{1}{50}$

=> $\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-..........-\frac{1}{7^{100}}<\frac{1}{50}$
 
You must log in or register to reply here.
Top Bottom