Question

 \(Cho\)\(A\) = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+......+\frac{1}{81}+\frac{1}{100}\)\(.\)\(CMR\)\(A\)> \(\frac{65}{132}\)

Answer 1

\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}+\frac{1}{10.10}\)

\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.10}\)

\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(A>1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

\(A>1+0+0+0+...+0-\frac{1}{10}\)

\(A>1-\frac{1}{10}=\frac{9}{10}\)

\(\Rightarrow A>\frac{5}{10}=\frac{1}{2}\)

mà : \(\frac{1}{2}=\frac{66}{132}>\frac{65}{132}\)

\(\Rightarrow A>\frac{65}{132}\)

Vậy \(A>\frac{65}{132}\)

Answer 2

phải là A = biểu thức đó và A = 9/10