Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
Bn tham khảo nhé :
2 / 2 . 3 + 2 /3 . 4 + 2 / 4 .5 + ... + 2 / 2017 . 2018
= 2 . ( 1/2 . 3 + 1/3 . 4 + 1/4 . 5 + ... + 1/ 2017 . 2018
= 2 . ( 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2017 - 1/2018 )
= 2 . ( 1/2 - 1/2018)
= 2 . 1008/2018
= 2016/2018
= 1008/1009
\(2\times(\frac{1}{2\times3}\times\frac{1}{3\times4}\times...\times\frac{1}{2017\times2018}))\)
\(2\times(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018})\)
\(2\times(\frac{1}{2}-\frac{1}{2018})\)
\(2\times\frac{504}{1009}=\frac{1008}{1009}\)
Bài làm:
Ta có: \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2008.2009}\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\right)\)
\(=2\left(1-\frac{1}{2009}\right)\)
\(=2.\frac{2008}{2009}=\frac{4016}{2009}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{2008.2009}=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2008.2009}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}\right)=2\left(1-\frac{1}{2009}\right)=2.\frac{2008}{2009}=\frac{4016}{2009}\)
a) Đặt A = 1.2 + 2.3 + ........ + (n-1)n
3A = 1.2.3 + 2.3.(4-1) + .... + (n-1)n[(n+1)-(n-2)]
3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + (n-1)n(n+1) - (n-2)(n-1)n
3A = (1.2.3 - 1.2..3) + ... + (n-1)n(n+1)
A = \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)
b) Đặt B = 12 + 22 + ..... + n2
B = 1(2 - 1) + 2(3 - 1) + ..... + n[(n + 1) - 1]
B = 1.2 + 2.3 + .......... + n(n + 1) - (1+2+3+....+n)
B = A - \(\frac{n\left(n+1\right)}{2}\)
\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\)
\(=\)\(\dfrac{1\cdot2}{2\cdot3}+\dfrac{1\cdot2}{3\cdot4}+...+\dfrac{1\cdot2}{99\cdot100}\)
\(=\)\(2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
\(=\)\(2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\)\(2\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=\)\(2\cdot\dfrac{49}{100}\)
\(=\)\(\dfrac{49}{50}\)
= 1/1 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1/1 - 1/100
= 99/100