(1 + 1 / 2) x (1 + 1/3 )x (1 + 1 / 4)....(1 + 1/ 99)
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= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10
= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10
=1-1/10=9/10
đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10
\(A=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+...+\frac{1}{9}\cdot\frac{1}{10}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100
\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2\times\frac{99}{100}\)
\(=\frac{99}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
vì tử của tất cả các số là 1-1 mà 1-1=0
suy ra:=0+0+0+...+0 (100 số 0)
Suy ra:=0
vậy (1-1/1+2).(1-1/1+2+3).....(1-1/1+2+3+...+99+100)=0
![](https://rs.olm.vn/images/avt/0.png?1311)
M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)
M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
M = \(1-\dfrac{1}{100}\)
M = \(\dfrac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
G = 1.2 + 1 + 2.3 + 1 + 3.4 + 1 + ... + 99.100 + 1
= 1.2 + 2.3 + 3.4 + ... + 99.100 + 99
⇒ 3G = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 99.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101 + 297
= 99.100.101 + 297
= 1000197
⇒ G = 1000197 : 3
= 333399
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\(M=1\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+...+\dfrac{1}{99}\times\dfrac{1}{100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\times....\times\left(1+\dfrac{1}{98}\right)\times\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{99}{98}\times\dfrac{100}{99}\)
\(=\dfrac{1\times100}{2\times1}\)
\(=\dfrac{100}{2}\\ =50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)......\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(=\left(-\dfrac{1}{2}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{3}{4}\right)......\left(-\dfrac{98}{99}\right)\left(-\dfrac{99}{100}\right)\)
\(=\dfrac{1}{100}\) ( do có 100 số hạng nên tích ra 1 số dương nhé ! ( chẵn mà )
Nhi có 10 viên bi,Nhi cho Oanh 5 viên bi.Hỏi Nhi còn lại bao nhiêu viên bi?
`(1+1/2) . (1+1/3) . (1+1/4)..... ( 1 + 1/99)`
`=(2/2+1/2) . (3/3 + 1/3) . (4/4+1/4) ..... ( 99/99 +1/99)`
`=3/2 . 4/3 . 5/4 ...... 100/99`
\(=\dfrac{3.4.5.6.....100}{2.3.4.....99}\)
`=100/2`
`=50`