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

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

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

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

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

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

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

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

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

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

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

sao anh đăng lắm câu hỏi vậy ?

 Đặt A=1/3 + 1/5 +1/7 +1/9+.....+1/243

      A=1/3 +(1/5+1/7+1/9)+(1/11+1/13+1/15+....+1/27)+(1/29+1/31+1/33+.......+1/81)+(1/83+1/85+1/87+...+1/243)

    => A>1/3+ 1/9 x3+1/27 x9+1/81x27+ 1/243x81=1/3x5=5/3

  => A>5/3>5/4

=>A>5/4

   => 1/3+1/5+1/7+.....+1/397 > 5/4

  =>1+1/3+1/5+1/7+.....+1/397 > 9/4

 =>1/5x (1+1/3+1/5+1/7+.....+1/397)> 9/4 x 1/5

    =>1/5+1/15+1/25+......+1/1985 > 9/20

??????????????????????????????????????????????????

sai đề rùi đáng lẽ > 9/20

Đề ??? :

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

Giải

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

\(\Rightarrow A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+...+\frac{1}{27}\right)+\left(\frac{1}{29}+...+\frac{1}{81}\right)+\left(\frac{1}{83}+...+\frac{1}{243}\right)\)

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

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

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

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

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

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

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

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

forever alone rai đề rùi