bn nhấn vào đúng 0 sẽ ra đáp án

2 uyên mắm

2 uyen mam

A=1/5+1/15+1/25+...+11/985
 A=1/5.(1+1/3+1.5+...+1/397)
=1/5.(1+1/1+2+1/2+3+...+1/198+199)
=1/5.(1+1−1/2+1/2−1/3+...+1/198−1/199)
=1/5.(2−1/199)
=397/995<920\

K nhé

Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+...+\frac{1}{243}\)

\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+...+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+...+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+...+\frac{1}{243}\right)\)

\(\Rightarrow A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81\)

\(=\frac{1}{3}.5\)

\(=\frac{5}{3}\)

\(\Rightarrow A>\frac{5}{3}>\frac{5}{4}\)

\(\Rightarrow A>\frac{5}{4}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{5}{4}\)

\(\Rightarrow1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{9}{4}\)

\(\Rightarrow\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}\right)>\frac{9}{4}.\frac{1}{5}\)

\(\Rightarrow\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}>\frac{9}{20}\)