Chứng tỏ rằng: 1/1.2+1/2.3+1/3.4+...+1/1999.2000<1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{1999}+\frac{1}{1999}\right)-\frac{1}{2000}\)
\(=\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)
\(=\frac{1}{1}-\frac{1}{2000}\)
\(=\frac{2000}{2000}-\frac{1}{2000}\)
\(=\frac{1999}{2000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/1999.2000
= 1 -1/2+1/2-1/3+1/3-1/4+....+1/1999-1/2000
= 1- 1/2000
= 1999/2000
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
Vì \(\frac{49}{50}
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)(TỐI GIẢN CÁC PHÂN SỐ LẬP LẠI )
\(A=\frac{99}{100}
Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
= \(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
= \(\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{100}{99.100}-\frac{99}{99.100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Vậy\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)< 1
~~~
#Sunrise
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)
Ta có: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)
=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}< 1\)
= \(\dfrac{1}{1}-\dfrac{1}{50}< 1\)
= \(\dfrac{50}{50}+\dfrac{-1}{50}< 1\)
= \(\dfrac{49}{50}< 1\)
Vậy \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phần chứng tỏ quy đồng lên rồi tính là ra
Còn phần tính S thì áp dụng tính chất vừa chứng tỏ để tách ra
Kết quả là 49/50
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\) (đpcm)
ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/1.2 + 1/2.3 + 1/3.4 + .......................+ 1/99.100
= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 +..................+ 1/99 - 1/100
= 1 - 1/100
= 99/100
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
Ma 99/100 < 1.
=> 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 < 1 (dccm)