Tính :
2/1.2 + 2/2.3 + 2/3.4 + ... + 2/99.100
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Mình làm mẫu 1 bài nha !
Có : 12A = 1.5.12+5.9.12+....+101.105.12
= 1.5.12+5.9.(13-1)+.....+101.105.(109-97)
= 1.5.12+5.9.13-1.5.9+.....+101.105.109-97.101.105
= 1.5.12-1.5.9+101.105.109
= 1155960
=> A = 1155960 : 12 = 96330
Tk mk nha
Có : 4D = 1.2.3.4+2.3.4.4+....+98.99.100.4
= 1.2.3.4+2.3.4.(5-1)+.....+98.99.100.(101-97)
= 1.2.3.4+2.3.4.5-1.2.3.4+......+98.99.100.101-97.98.99.100
= 98.99.100.101
=> D = 98.99.100.101/4 = 24497550
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{98.99}+\frac{2}{99.100}\)
= \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
= \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
= \(2\left(1-\frac{1}{100}\right)\)
=\(2.\frac{99}{100}\)
=\(\frac{99}{50}\)
= 2/1 - 2/2 + 2/2 - 2/3 + 2/3 - 2/4 + ..... + 2/99 - 2/100
= 2/1 + 2/100
= 101/50
ta có
\(A=1.2^2+2.3^2+3.4^2+...+99.100^2=1.2.\left(3-1\right)+2.3.\left(4-2\right)+...+99.100.\left(101-99\right)\)
\(A=\left(1.2.3+2.3.4+...+99.100.101\right)-\left(2.3+3.4+...+99.100\right)\)Đối với bt trước ông nhân với 4 =>đc tổng 98.99.100.101
Đối với bt sau ông nhân với 3 được tổng là 99.100.101
=>A=98.99.100.101 - 99.100.101=97.99.100.101=96990300
nhớ tick nha lắc lư
A = 1.22 + 2.32 + 3.42 + …. + 99.1002
A= 1.2.2 + 2.3.3 + 3.4.4 +...+99.100.100
A= 1.2(3-1) +2.3(4-1) +3.4(5-1) +....+ 99.100(101-1)
A= 1.2.3 - 1.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 1.3.4 +...+99.100.101- 1.99.100
A= 1.2.3 + 2.3.4 + 3.4.5+....+99.100.101 - 1.2 +2.3 + 3.4+...+ 99.100
A= 24497550 - 333300
A=24164250
Vậy...
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{99.100}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}=\frac{99}{50}\)
=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
=\(2\left(1-\frac{1}{100}\right)\)
=\(2\cdot\frac{99}{100}=\frac{99}{50}\)
mk k vt lại đề nha
S=2.(1/1.2+1/2.3+1/3.4+............+1/99.100)
S=2.(1-1/2+1/3-1/4+1/4-1/5+.............+1/99-1/100)
S=2.(1-1/100)
S=2.99/100
S=198/100
S=\(\frac{2}{1.2}\)+\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)+...+\(\frac{2}{98.99}\)+\(\frac{2}{99.100}\)
S=\(\frac{2}{1}\).(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{98.99}\)+\(\frac{1}{99.100}\))
S=\(\frac{2}{1}\).(\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{98}\)-\(\frac{1}{99}\)+\(\frac{1}{99}\)-\(\frac{1}{100}\))
S=\(\frac{2}{1}\).(\(\frac{1}{1}\)-\(\frac{1}{100}\))
S=\(\frac{2}{1}\).(\(\frac{100}{100}\)-\(\frac{1}{100}\))
S=\(\frac{2}{1}\).\(\frac{99}{100}\)
S=\(\frac{99}{50}\)
Vậy S=\(\frac{99}{50}\)
\(P=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\\ =2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\\ =2\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\cdot\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =2\cdot\dfrac{99}{100}\\ =\dfrac{99}{50}\)
\(P=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\\ =2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\\ =2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)
ta có 1+(1+2)+(1+2+3)+...+(1+2+3+...+100)
=4+(1+3).3/2+9+(1+4).4/2+...+(1+100).100/2
=1/2(1.2+2.3+.....+100.101)
=>1/2.100.101.102
con cái dưới thì bằng 99.100.101
=>F=51/99
ngu rua mà ko biet lam
\(= 1.(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3}+\frac{1}{3} - ... + \frac{1}{99} + \frac{1}{100})\)
\(= 1. ( 1-\frac{1}{100}) \)
\(= \frac{99}{100}\)
*Bài trước nhân 1 là sai mình sửa*
\(=2. (\frac{1}{1.2} + ..+ \frac{1}{99.100}) \)
\(= 2.(\frac{1}{1} - \frac{1}{2} + \frac{1}{2}-\frac{1}{3}+\frac{1}{3} - ... + \frac{1}{99} + \frac{1}{100})\)
\(= 2.(1-\frac{1}{99}) = \frac{99}{100} . 2 = \frac{99}{50}\)