Question

[Nguyễn Thị Thu HIền]

cho A = 1/10 + 1/11 + 1/12+ ...+ 1/99 + 1/ 100

chứng Minh Rằng A > 1

Answer 1

[Nguyễn Thanh Hằng]

Ta có :

A = \(\dfrac{1}{10}\) + \(\dfrac{1}{11}\) + \(\dfrac{1}{12}\) +.................+ \(\dfrac{1}{99}\) + \(\dfrac{1}{100}\) ( 91 số hạng)

A = \(\dfrac{1}{10}\) + \(\left(\dfrac{1}{11}+\dfrac{1}{12}+...........+\dfrac{1}{99}+\dfrac{1}{100}\right)\)

Vì \(\dfrac{1}{11}>\dfrac{1}{100}\)

\(\dfrac{1}{12}>\dfrac{1}{100}\)

.................................

\(\dfrac{1}{99}< \dfrac{1}{100}\)

\(=>\) \(A\) > \(\dfrac{1}{10}+\left(\dfrac{1}{100}+\dfrac{1}{100}+........+\dfrac{1}{100}\right)\) (90 số hạng \(\dfrac{1}{100}\) )

A > \(\dfrac{1}{10}+\dfrac{90}{100}\)

\(A\) > \(\dfrac{1}{10}+\dfrac{9}{10}\)

=> A > 1

=> đpcm