chứng tỏ rằng : 1/5 + 1/15 + 1/25 + ......+1/1985 < 9/20
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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 số gần bằng) => A= 1,2124 (lấy số gần bằng)
=> A < 9/20
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}\)
\(A=\dfrac{1}{5}.\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{397}\right)\)
\(A=\dfrac{1}{5}.\left(1+\dfrac{1}{1+2}+\dfrac{1}{2+3}+...+\dfrac{1}{198+199}\right)\)
\(A=\dfrac{1}{5}.\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{198}-\dfrac{1}{199}\right)\)
\(A=\dfrac{1}{5}.\left(2-\dfrac{1}{199}\right)\)
\(A=\dfrac{397}{995}< \dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}< \dfrac{9}{20}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Nhân 2 vế với 5 suy ra
1+1/3+1/5+...+1/397<9/4 hay
1/3+1/5+...+1/397<5/4
Trong khi tính thử bằng máy tính thì
1/3+1/5+...+1/397>1/3+1/5+1/7+...+1/25 = 1,2643528....>5/4
tích nha