Cho A=1/1.3 +1/3.5+1/5.7 +... +1/99.100
Khi đó 200A = ?
Cho A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+........+\frac{1}{99.100}\)
khi đó 200A bằng ....
Cho A= \(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+...+\(\frac{1}{99.100}\). Khi đó 200A=
a) 1/1.3+1/3.5+1/5.7
b) 1/1.3+1/3.5+1/5.7+...+1/2007.2009+1/2009.2011
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}\)
\(=\frac{3}{7}\)
b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)
Cho S=1/1.3+1/3.5+1/5.7+...+1/99.100. Khi đó 2S+1/101
có dạng này nhưng là số chẵn nhân chãn
Cho S = 1/1.3 + 1/3.5 +1/5.7 + ......+ 1/99.101. Khi đó 2S + 1/101 = ?
2S=2/1.3+2/3.5+....+2/99.101
2S=1-1/3+1/3-1/5+....+1/99-1/101
2S=1-1/101
2S+1/101=1-1/101+1/101=1
Nho tick nha
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(S=1-\frac{1}{101}=\frac{100}{101}\)
\(2S+\frac{1}{101}=\frac{100}{101}\)
\(S=2.\frac{100}{101}+\frac{1}{101}\)
\(\Rightarrow S=\frac{201}{101}\)
****
2S + \(\frac{1}{101}\)=\(\frac{201}{101}\)
Cho S=1/1.3+1/3.5+1/5.7+............+1/99.101
Khi đó 2S+1/101=...........
Trả lời :............
A = 1/1.3 - 1/3.5 - 1/5.7 - ... - 1/99.101
\(=\dfrac{1}{3}-\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{98}{303}=\dfrac{1}{3}-\dfrac{49}{303}=\dfrac{101-49}{303}=\dfrac{52}{303}\)
Cho biểu thức A=1/1.3+1/3.5+1/5.7+...+1/37.39.Hãy so sánh A với 1/2
\(A=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{37\cdot39}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{38}{39}< \dfrac{1}{2}\)
a)Tính: 1/1.3 + 1/3.5 + 1/5.7 +...+1/19.21
b) chúng minh: A= 1/1.3 + 1/3.5 +...+ 1/(2n-1)(2n+1) < 1/2
a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21
Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21
= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21
= 1-1/21 = 20/21
=> B= 20/21 : 2 => B= 10/21
b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2
=> A= 1/2- 1/2n+1
=> A< 1/2 ( đpcm )