Tính :B=2/1x2+2/2x3+2/3x4……+2/99x100
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HV
0
BT
11 tháng 3 2017
A=2(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\))=2(\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\))
=> A=2(\(\frac{1}{1}-\frac{1}{100}\))=2.\(\frac{99}{100}=\frac{99}{50}\)
ĐS: A=99/50
11 tháng 3 2017
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{99\times100}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{99}{100}\)
NA
1
HT
27 tháng 10 2015
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{99.100}\)
= \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
= \(2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)
= \(2.\left(1-\frac{1}{100}\right)\)
= \(2.\frac{99}{100}\)
= \(\frac{99}{50}\)
HV
3
15 tháng 1 2022
\(\Leftrightarrow y\cdot\dfrac{99}{50}=\dfrac{198}{100}=\dfrac{99}{50}\)
hay y=1
15 tháng 7 2017
Ta có : P = 1.2.2 + 2.3.3 + ....+ 99.100.100
=1.2.(3 - 1) + 2.3.(4 - 1) + ....+99.100.(101 - 1)
= (1.2.3 + 2.3.4 + .... + 99.100.101) - (2.3 + 3.4+.....+99.100)
Đặt B = 1.2.3 + 2.3.4 + 4.5.6 +...+ 99.100.101
4B = 1.2.3.(4 - 0)+2.3.4.(5 - 1) + ... + (99.100.101(102 - 98)
4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 +...+ 99.100.101.102 - 98.99.100.101
4B = 99.100.101.102
4B = 101989800
B = 25497450
Đặt C = 1.2 + 2.3 + 3.4 +...+ 99.100
3C = 1.2.(3 - 0) + 2.3.(4 - 1) +...+ 99.100.(101 - 98)
3C = 1.2.3 + 2.3.4 - 1.2.3 +...+ 99.100.101 - 98.99.100
3C = 99.100.101
3C = 999900
C = 999900 : 3
C = 333300
Vậy: P = 25497450 – 333300 = 25164150
NT
2
10 tháng 9 2017
Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{99.100}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=2.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2.99}{100}\)
\(A=\frac{99}{50}=1\frac{49}{50}\)
10 tháng 9 2017
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)=2.\frac{99}{100}\)
\(=\frac{99}{50}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 7 2023
1x 2 + 2 x 3 + 3 x 4 + ...+ 99 x 100
Ta có:
1 x 2 x 3 = 1 x 2 x 3
2 x 3 x 3 = 2 x 3 x ( 4 - 1) = 2 x 3 x 4 - 1 x 2 x 3
3 x 4 x 3 = 3 x 4 x ( 5 - 2) = 3 x 4 x 5 - 2 x 3 x 4
........................................................= ........................................
99 x 100 x 3 = 99 x 100 x (101 - 98) = 99 x 100 x 101 - 99 x 100 x 98
Cộng vế với vế ta có:
1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 +...+ 99 x 100 x 3 = 99 x100 x 101
(1 x 2 + 2 x 3 + 3 x 4 +...+ 99 x 100) x 3 = 99 x 100 x 101
1 x 2 + 2 x 3 + 3 x 4 +...+ 99 x 100 = \(\dfrac{99\times100\times101}{3}\)
1 x 2 + 2 x 3 + 3 x 4 + ....+ 99 x 100 = 333300
27 tháng 1 2017
A = 1.2+2.3+3.4+......+99.100
Gấp A lên 3 lần ta có:
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100
A . 3 = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
A = 333 300
PT
1
CM
10 tháng 11 2017
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3 A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) ..................................
A x 3 = 99x100x101 A = 333300
B = \(\dfrac{2}{1\times2}\) + \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+ \(\dfrac{2}{99\times100}\)
B = 2 \(\times\) ( \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+....+ \(\dfrac{1}{99\times100}\))
B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+...+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\))
B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))
B = 2 \(\times\) \(\dfrac{99}{100}\)
B = \(\dfrac{99}{50}\)