B=1.2+2.3+......+199.200
Tính B
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34
=> 3S = 32.33.34
=> S = \(\frac{32.33.34}{3}=11968\)
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
Ta có : B = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
=> 3B = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
=>3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
=> 3B = 99.100.101
=> 3B = 999900
=> B = 333300
Vậy B = 333300
Bài làm :
Ta có :
B= 1.2 + 2.3 + 3.4 + ...+ 99.100
=>3B = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
<=>3B= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
<=>3B= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
<=>3S = 99.100.101
<=> 3S = 999900
<=> B = 999900 : 3 = 333300
Vậy B = 333300
Đặt \(A=1.2+2.3+3.4+...+2015.2016\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2015.2016.\left(2017-2014\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016\)
\(\Rightarrow3A=2015.2016.2017\)
\(\Rightarrow A=2015.2016.2017:3\)
\(\Rightarrow A=2015.672.2017\)
Vậy \(A=2015.672.2017\)
1 . 2 + 2 . 3 + 3 . 4 + ... + 2015 . 2016
3M = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 2015 . 2016 . 3
3M = 1 . 2 ( 3 - 0 ) + 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 2015 . 2016 ( 2017 - 2014 )
3M = ( 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4. 5 + ... + 2015 . 2016 . 2017 ) - ( 0 . 1 . 2 + 1 . 2 . 3 + 2 . 3 . 4 + ... + 2014 . 2015 . 2016 )
3M = 2015 . 2016 . 2017
M = \(\frac{2015.2016.2017}{3}\)
M = 2731179360