5/6 + 11/12 + 19/20 + ...+ 9701/9702 + 9899/9900
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VY
1
5 tháng 5 2018
Có: \(A=\frac{1}{2}+\frac{5}{6}+...+\frac{9899}{9900}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{9900}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)\)
\(=99-\frac{99}{100}< 99\)
\(\Rightarrow A< 99\)
ND
1
12 tháng 8 2016
1/2 + 5/6 + 11/12 + 19/20 + 29/30 +. . . 9701 + / 9702 + 9899/9900 = 1/2 + (1-1 / 6) + (1-1 / 12) + (1-1 / 20) + (1-1 / 30) + ...... + (1 -1/9702) + (1-1 / 9900) = 1/2 + [1 - (1 / 2-1 / 3)] + [1 - (1 / 3-1 / 4)] + [1- ( 1 / 4-1 / 5)] + [1 - (1 / 5-1 / 6)] + ...... + [1- (1 / 98-1 / 99)] + [1 - (1 / 99-1 / 100)] * 100 + 1 = 1 / 2-1 / 2 + 1 / 3-1 / 3 + 1 / 4-1 / 4 + 1 / 5-1 / 5 + 1 / 6-1 / 6 + ... ... 1 / 98-1 / 98 + 1 / 99-1 / 99 + 1/100 + 1 = 100/100 = 100 và 1/100
CN
1
MV
4 tháng 5 2017
\(A=100\cdot\left(1+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+...+\dfrac{9899}{9900}\right)\\ =100\cdot\left(1+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+...+1-\dfrac{1}{9900}\right)\\ =100\cdot\left[\left(1+1+1+...+1\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\right]\\ =100\cdot\left[99-\dfrac{49}{100}\right]\\ =100\cdot\dfrac{9851}{100}\\ =9851\)
25 tháng 4 2018
Ta có :
\(A=100\left(1+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\right)\)
\(A=100\left(1+\frac{6-1}{6}+\frac{12-1}{12}+\frac{20-1}{20}+...+\frac{9900-1}{9900}\right)\)
\(A=100\left(1+\frac{6}{6}-\frac{1}{6}+\frac{12}{12}-\frac{1}{12}+\frac{20}{20}-\frac{1}{20}+...+\frac{9900}{9900}-\frac{1}{9900}\right)\)
\(A=100\left(1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\right)\)
\(\frac{A}{100}=1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{100}\right)\)
Do từ \(2\) đến \(99\) có \(99-2+1=98\) số nên có \(98\) số \(1\) suy ra :
\(\frac{A}{100}=98-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\frac{A}{100}=98-\frac{49}{100}\)
\(\frac{A}{100}=\frac{9751}{100}\)
\(A=\frac{9751}{100}.100\)
\(A=9751\)
Vậy \(A=9751\)
Chúc bạn học tốt ~
1 tháng 4 2018
D=\(1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+........+1-\frac{1}{9900}\)
\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)\)
\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=98+\frac{1}{100}=\frac{9801}{100}\)
1 tháng 4 2018
d=1/1.2+5/2.3+11/3.4+...+9899/99.100
=>d=1-1/2+1/2-1/3+...+1/99-1/100
=>d=1-1/100
=>d=99/100
Vậy d=99/100
27 tháng 2 2016
A= 2 + 6 + 12 + 20 + ...... + 9702 + 9900
A = 1.2 + 2.3 + 3.4 + ......... + 98 . 99 + 99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3
3A = 1.2.3 + 2.3.(4-1) + ....+ 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ..... + 99.100.101 - 98.99.100
3A = 99 . 100 . 101
A = 99 . 100 . 101 : 3
A = 333300
27 tháng 2 2016
A=1.2+2.3+3.4+4.5+....+.....
3A=.....
Bạn biets làm rồi đúng ko
Tích mk nha hùng
\(\dfrac{5}{6}+\dfrac{11}{12}+...+\dfrac{9701}{9702}+\dfrac{9899}{9900}\)
\(=1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{9702}+1-\dfrac{1}{9900}\)
\(=1-\dfrac{1}{2\cdot3}+1-\dfrac{1}{3\cdot4}+...+1-\dfrac{1}{99\cdot100}\)
\(=98-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=98-\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=98-\dfrac{49}{100}=97,51\)