Cho tổng A=1/2+5/6+11/12+19/20+...+9701/9702+9899/9900
Chứng tỏ A<99
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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 ~
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
BN
2
16 tháng 5 2022
`= 1 - 1/2 + 1 - 1/6 + ... + 1 - 1/56`
`= 1 - 1/(1.2) + 1 - 1/(2.3) + ... + 1 - 1/(7.8)`
`= 7 - (1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8`.
`= 8 - 1/8`
`= 63/64`.
2
16 tháng 5 2022
`A=1/2+5/6+11/12+19/20+29/30+41/42+55/56`
`A=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56`
`A=(1+1+1+1+1+1+1)-(1/2+1/6+1/12+....+1/56)`
`A=7-(1/[1xx2]+1/[2xx3]+1/[3xx4]+....+1/[7xx8])`
`A=7-(1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8)`
`A=7-(1-1/8)`
`A=7-(8/8-1/8)`
`A=7-7/8`
`A=56/8-7/8=49/8`
NH
1
6 tháng 8 2016
A = 1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72
A = ( 1 - 1/2 ) + ( 1 - 1/6 ) + ( 1 - 1/12 ) + ( 1 - 1/20 ) + ( 1 - 1/30 ) + ( 1 - 1/42 ) + ( 1 - 1/56 ) + ( 1 - 1/72 )
A = 1 x 8 - ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 )
A = 8 - ( \(\frac{1}{1\cdot2} +\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\))
A = \(8-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
A = \(8-\left(1-\frac{1}{9}\right)\)
\(A=8-\frac{8}{9}\)
\(A=\frac{64}{9}\)
Có: \(A=\frac{1}{2}+\frac{5}{6}+...+\frac{9899}{9900}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{9900}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)\)
\(=99-\frac{99}{100}< 99\)
\(\Rightarrow A< 99\)