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A=1/8+1/24+1/48+1/80+...+1/360
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HN
1
LM
5 tháng 5 2022
2a= 2/3+2/8+2/15+2/24+2/35+2/48+2/63+2/80= [2/( 1*3)+2/( 3*5)+2/( 5*7)+2/( 7*9)]+[2/(2*4)+2/(4*6)+2/(6*8)+2/(8*10)]= [1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9]+[1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10]= [1/1-1/9]+[1/2-1/10]= 8/9+2/5= 58/45 =>a= 29/45
2 tháng 3 2023
\(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(=1+\dfrac{1}{2\times4}+\dfrac{1}{4\times6}+\dfrac{1}{6\times8}+\dfrac{1}{8\times10}+\dfrac{1}{10\times12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{17}{12}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 11 2023
8 = 2 \(\times\) 4
24 = 4 \(\times\) 6
48 = 6 \(\times\) 8
80 = 8 \(\times\) 10
Xét dãy số: 2; 4; 6; 8;...; đây là dãy số cách đều với khoảng cách là:
4 - 2 = 2
Số thứ 20 của dãy số trên là: 2 x (20 - 1) + 2 = 40
Vậy Phân số thứ 20 của dãy số đã cho là: \(\dfrac{1}{40\times42}\)
Tổng của 20 phân số đầu tiên của dãy số đã cho là:
A = \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{48}\) + \(\dfrac{1}{80}\) +...+ \(\dfrac{1}{1680}\)
A = \(\dfrac{1}{2\times4}\) + \(\dfrac{1}{4\times6}\) + \(\dfrac{1}{6\times8}\) + \(\dfrac{1}{8\times10}\)+...+ \(\dfrac{1}{40\times42}\)
A = \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{2\times4}\) + \(\dfrac{2}{4\times6}\)+\(\dfrac{2}{6\times8}\)+\(\dfrac{2}{8\times10}\)+...+\(\dfrac{2}{40\times42}\))
A = \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{10}\)+...+ \(\dfrac{1}{40}\) - \(\dfrac{1}{42}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{42}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{40}{42}\)
A = \(\dfrac{5}{21}\)
28 tháng 2 2018
A = 1 + 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10 + 1/10.12
2A = 2 + 2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 + 2/10.12
= 2 + 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + 1/8 - 1/10 + 1/10 - 1/12
= 2 + 1/2 - 1/12 = 29/12
=> A = 29/12 : 2 = 29/24
Tk mk nha
NT
3
TT
5 tháng 5 2022
A=1+18+124+148+180+1120A=1+18+124+148+180+1120
=1+12.4+14.6+16.8+18.10+110.12=1+12.4+14.6+16.8+18.10+110.12
=1+12(12−14+14−16+16−18+18−110+110−112)=1+12(12−14+14−16+16−18+18−110+110−112)
=1+12(12−112)=1+12(12−112)
=1+524=1+524
=2924
Tham khảo thôi nka
LM
5 tháng 5 2022
2A= 2/8+2/24+2/48+2/80= 2/(2*4)+2/(4*6)+2/(6*8)+2/(8*10)= 1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10= 1/2-1/10= 2/5 =>A= 1/5
TD
1
5 tháng 4 2021
Ta có: \(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(\Leftrightarrow2A=2+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{24}{12}+\dfrac{6}{12}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{29}{12}\)
hay \(A=\dfrac{29}{24}\)
YA
3
9 tháng 2 2018
A = 1 + 1/2.4 + 1/4.6 + ...... + 1/10.12
2A = 2 + 2/2.4 + 2/4.6 + ...... + 2/10.12
= 2 + 1/2 - 1/4 + 1/4 - 1/6 + ...... + 1/10 - 1/12
= 2 + 1/2 - 1/12 = 29/12
=> A = 29/12 : 2 = 29/24
P/S : Tham khảo nha
12 tháng 2 2018
\(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{440}\)
\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+....+\frac{1}{20\cdot22}\)
\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+.....+\frac{2}{20\cdot22}\)
\(2A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{20}-\frac{1}{22}\)
\(2A=1-\frac{1}{22}\)
\(A=\frac{21}{22}:2\)
\(A=\frac{21}{44}\)
12 tháng 2 2018
\(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+...+\frac{1}{440}\)
= \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{20.22}\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{20}-\frac{1}{22}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{22}\right)=\frac{1}{2}.\frac{5}{11}=\frac{5}{22}\)
TL
0
\(A=\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+...+\dfrac{1}{360}\)
\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+...+\dfrac{1}{18\cdot20}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{18\cdot20}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{2}\cdot\dfrac{9}{20}=\dfrac{9}{40}\)