tính
A=1.4+4.7+7.10+.......+2014.2017
K
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Những câu hỏi liên quan
B
3
G
4 tháng 2 2020
A = \(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+ \(\frac{1}{2014.2017}\)
3A = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2014.2017}\)
3A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{2014}-\frac{1}{2017}\)
3A= 1 - \(\frac{1}{2017}\)
A = \(\frac{1}{3}-\frac{1}{2017.3}\)
A = \(\frac{672}{2017}\)
YN
4 tháng 2 2020
Ta có \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2014.2017}\)
\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{3}.\frac{2016}{2017}=\frac{672}{2017}\)
Vậy \(A=\frac{672}{2017}\)
~ Học tốt
# Chiyuki Fujito
3 tháng 3 2017
\(B=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+...+\dfrac{4}{2014\cdot2017}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2014\cdot2017}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{2017}\right)=\dfrac{4}{3}\cdot\dfrac{2016}{2017}=\dfrac{8064}{6051}\)
22 tháng 10 2018
\(a,A=\dfrac{101}{100}+\dfrac{102}{100}+\dfrac{103}{100}+...+\dfrac{199}{100}\)
\(A=\dfrac{101+102+103+...+109}{100}\)
Xét tử số : \(101+102+103+...+199\)
Có : \(\left(199-101\right):1+1=99\) (số hạng)
\(\Rightarrow\) Tử số bằng \(:\left(199+101\right).99:2=14850\)
\(\Rightarrow A=\dfrac{14850}{100}=\dfrac{297}{2}\)
\(b,B=\dfrac{10002}{10000}+\dfrac{10004}{10000}+\dfrac{10006}{10000}+...+\dfrac{12014}{10000}\)
\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)
\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)
Xét tử số : \(10002+10004+10006+...+12014\)
Có : \(\left(12014-10002\right):2+1=1007\) (số hạng)
\(\Rightarrow\) Tử số bằng : \(\left(12014+10002\right).1007:2=11085056\)
\(\Rightarrow B=\dfrac{11085056}{10000}\)
Bạn tự làm câu C nha
\(D=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)
\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{2015}=\dfrac{402}{2015}\)
\(E=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2014.2017}\)
\(\Rightarrow3E=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\)
\(\Rightarrow3E=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\)
\(\Rightarrow3E=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)
\(\Rightarrow E=\dfrac{2016}{2017}:3=\dfrac{672}{2017}\)
22 tháng 10 2018
D = \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) +...+ \(\dfrac{1}{2014.2015}\)
D = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)+...+ \(\dfrac{1}{2014}-\dfrac{1}{2015}\)
D = \(\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
D = \(\dfrac{403}{2015}-\dfrac{1}{2015}\)
D = \(\dfrac{402}{2015}\)
BB
8
3 tháng 3 2017
Ta có : B = \(\frac{4}{1.4}+\frac{4}{4.7}+\frac{4}{7.10}+......+\frac{4}{2014.2017}\)
\(=\frac{4}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{2014.2017}\right)\)
\(=\frac{4}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+......+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(=\frac{4}{3}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{4}{3}.\frac{2016}{2017}=\frac{2688}{2017}\)
2 tháng 10 2023
`#3107.101107`
1.
a)
`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`
`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`
`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`
`= 1/3* (1 - 1/103)`
`= 1/3*102/103`
`= 34/103`
b)
`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`
`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`
`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`
`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`
`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`
`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`
`= -1/2 * (1 - 1/101)`
`= -1/2*100/101`
`= -50/101`
2.
`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`
`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`
`= 1-1/100`
`= 99/100`
30 tháng 4 2019
Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{2014\cdot2017}\)
\(\Rightarrow A=\frac{1}{3}\cdot\left(\frac{3}{1\cdot3}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{2014\cdot2017}\right)\)
\(\Rightarrow A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{3}\cdot\left(1-\frac{1}{2017}\right)=\frac{1}{3}-\frac{1}{6051}< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{3}\left(ĐPCM\right)\)
KN
30 tháng 4 2019
Ta có :
\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2014.2017}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{3}.\frac{2016}{2017}< \frac{1}{3}\left(đpcm\right)\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
S = 1.4 + 4.7 + 7.10 + 10.13 + ... + 61.64
1.4.9 = 1.4.(7 + 2) = 1.4.7 + 1.4.2
4.7.9 = 4.7.(10 - 1) = 4.7.10 - 1.4.7
7.10.9 = 7.10.(13 - 4) = 7.10.13 - 4.7.10
10.13.9 = 10.13.(16 - 7) = 10.13.16 - 7.10.13
.......................................................................
61.64.9 = 61.64.(67 - 58) = 61.64.67 - 58.61.64
Cộng vế với vế ta có:
1.4.9 + 4.7.9 + 7.10.9 +...+ 61.64.9 = 1.4.2 + 61.64.67
9(1.4 + 4.7 + 7.10+ ...+ 61.64) = 261576
1.4 + 4.7 + 7.10 +...+ 61.64 = 261576 : 9
1.4 + 4.7 + 7.10 + ... + 61.64 = 29064
K
11 tháng 6 2019
\(\text{Đặt: S= biểu thức cần tính}\)
\(\Rightarrow9S=1.4.7+4.7.9+......+19.22.9+4.2\)
\(\Rightarrow9S=1.4.7+4.7\left(10-1\right)+...+19.22\left(25-16\right)+8\)
\(\Rightarrow9S=19.22.25+8\Rightarrow S=1162\)
NT
4
10 tháng 5 2017
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)
Q
10 tháng 5 2017
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)
\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)
\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(3M=5\left(1-\frac{1}{2017}\right)\)
\(3M=5.\frac{2016}{2017}\)
\(3M=\frac{10080}{2017}\)
\(\Rightarrow M=\frac{3360}{2017}\)
