Chứng minh rằng:
\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+..........+\frac{1}{1985}< \frac{9}{20}\)
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NH
1
PT
16 tháng 1 2017
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+...+\frac{1}{243}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+...+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+...+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+...+\frac{1}{243}\right)\)
\(\Rightarrow A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81\)
\(=\frac{1}{3}.5\)
\(=\frac{5}{3}\)
\(\Rightarrow A>\frac{5}{3}>\frac{5}{4}\)
\(\Rightarrow A>\frac{5}{4}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{5}{4}\)
\(\Rightarrow1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{9}{4}\)
\(\Rightarrow\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}\right)>\frac{9}{4}.\frac{1}{5}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}>\frac{9}{20}\)
27 tháng 1 2016
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
KN
16 tháng 4 2019
Đề ??? :
\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}>\frac{9}{20}\)
Giải
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{243}\)
\(\Rightarrow A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+...+\frac{1}{27}\right)+\left(\frac{1}{29}+...+\frac{1}{81}\right)+\left(\frac{1}{83}+...+\frac{1}{243}\right)\)
\(\Rightarrow A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81\)
\(=\frac{1}{3}.5\)
\(=\frac{5}{3}\)
\(\Rightarrow A>\frac{5}{3}>\frac{5}{4}\)
\(\Rightarrow A>\frac{5}{4}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{5}+...+\frac{1}{397}>\frac{9}{4}\)
\(\Rightarrow\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{397}\right)>\frac{9}{4}.\frac{1}{5}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}>\frac{9}{20}\left(đpcm\right)\)
L
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NV
1
NT
0