Tính 1/1x3 + 1/3x5 + 1/5x7+....+1/19x21
Chứng minh A=1/1x3+1/3x5+.....+1/(2n-1)x(2n+1)<1/2
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A
1
9 tháng 7 2016
1/1x3 + 1/3x5 + 1/5x7 + ... + 1/(2n+1)x(2n+3) = n+1/2n+3
2/1x3 + 2/3x5 + 2/5x7 + ... + 2/(2n+1)x(2n+3) = 2n+2/2n+3
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2n+1 - 1/2n+3 = 2n+2/2n+3
1 - 1/2n+3 = 2n+2/2n+3
Bn nào thông minh thế, ra bài này đố Tây lm đc, ai lm đc mk bái lm sư phụ lun, sửa đề đê
Ủng hộ mk nha ^_-
DB
3
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
AH
3
11 tháng 5 2017
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\frac{2016}{2017}\)
\(S=\frac{1008}{2017}< \frac{1}{2}\)
S
11 tháng 5 2017
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(2S=1-\frac{1}{2017}< 1\)
=> 2S < 1
=> S < \(\frac{1}{2}\)(đpcm)
VT
1
TM
10 tháng 7 2016
\(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{\left(2n+1\right)x\left(2x+3\right)}=\frac{n+1}{2n+3}\)
=>\(2x\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{\left(2n+1\right)x\left(2n+3\right)}\right)=2x\frac{n+1}{2n+3}\)
=>\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}=\frac{2n+2}{2n+3}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)
=>\(1-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)
=>\(\frac{2n+2}{2n+3}=\frac{2n+2}{2n+3}\)
=>.....
AH
Akai Haruma
Giáo viên
31 tháng 12 2022
Lời giải:
$2\times A=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{19\times 21}$
$2\times A=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+...+\frac{21-19}{19\times 21}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{21}$
$=1-\frac{1}{21}=\frac{20}{21}$
$\Rightarrow A=\frac{20}{21}: 2= \frac{10}{21}$
HT
27 tháng 10 2020
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
= \(1-\dfrac{1}{101}\)
=\(\dfrac{100}{101}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)
=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
=\(\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}-\dfrac{100}{101}\)
= \(\dfrac{305}{202}\)
a, Đặt :
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..............+\dfrac{1}{19.21}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{19.21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{19}-\dfrac{1}{21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{21}\)
\(\Leftrightarrow2A=\dfrac{20}{21}\)
\(\Leftrightarrow A=\dfrac{10}{21}\)
b, \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...........+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
\(\Leftrightarrow2A=1-\dfrac{1}{2n+1}\)
\(\Leftrightarrow2A=\dfrac{2n}{2n+1}\)
\(\Leftrightarrow A=\dfrac{n}{2n+1}\)