trợ giúp

H

hiensau99

Mình viết lại đề: Chứng Minh Rằng: [TEX]\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{199}{99^2.100^2}<1[/TEX]


$\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+ \frac{197}{98^2.99^2}+\frac{199}{99^2.100^2}$


$= \frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+ \frac{99^2-98^2}{98^2.99^2} + \frac{100^2-99^2}{99^2.100^2}$

$= \frac{2^2}{1^2.2^2} - \frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+... + \frac{99^2}{98^2.99^2}- \frac{98^2}{98^2.99^2}+\frac{100^2}{99^2.100^2} - \frac{99^2}{99^2.100^2}$

$= 1 - \frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+ \frac{1}{98^2}- \frac{1}{99^2}+ \frac{1}{99^2} - \frac{1}{100^2}$

$= 1 - \frac{1}{100^2} < 1$ (đpcm)


 
Top Bottom