CMR \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}<\frac{9}{20}\)
Cho xin lời giải nhá mọi người
CMR:\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}< \frac{9}{20}\)
CMR:
\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}< \frac{9}{20}\)
CMR \(A=\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}< \frac{9}{20}\)
CMR: \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}<\frac{9}{20}\)
CMR \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\) +..................+\(\frac{1}{1985}<\frac{9}{20}\)
Cho : A= \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}\)
CMR: A<\(\frac{9}{20}\)
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
CMR: \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{1985}<\frac{9}{20}\)
CMR:
\(y=\frac{1}{5}\)\(+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}\)< \(\frac{9}{20}\)
A = (1/5)+(1/15)+(1/25)+...+(1/1985)=
1/5+1/3*5+1/5*5+1/7*5+.........+1/397*5
5A=1+1/3+1/5+1/7+.......+1/397
5A-1=1/3+1/5+1/7+.......+1/397
Đặt B= 1/3+1/5+1/7+.......+1/397
=>.......................Tính đc B=5,06241 (lấy gần bằng)=> A= 1,2124 (lấy số gần bằng)
=> A < 9/20
Chứng minh rằng: \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+..........+\frac{1}{1985}< \frac{9}{20}\)
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+...+\frac{1}{243}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+...+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+...+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+...+\frac{1}{243}\right)\)
\(\Rightarrow A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81\)
\(=\frac{1}{3}.5\)
\(=\frac{5}{3}\)
\(\Rightarrow A>\frac{5}{3}>\frac{5}{4}\)
\(\Rightarrow A>\frac{5}{4}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{5}{4}\)
\(\Rightarrow1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}>\frac{9}{4}\)
\(\Rightarrow\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{397}\right)>\frac{9}{4}.\frac{1}{5}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+...+\frac{1}{1985}>\frac{9}{20}\)