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số số hạng của A là :
( 2007 - 3 ) : 3 + 1 = 669 ( số )
tổng A là :
( 2007 + 3 ) . 669 : 2 = 672345
B = \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...+\left(\dfrac{1}{2006}+1\right)+1}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}+\dfrac{2007}{2007}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{2007.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2006}+\dfrac{1}{2007}\right)}\)
B = \(\dfrac{2006}{2007}\)
2006/1 là 2006, tách 1 của 2006 ra 2005 phân số còn lại 1
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A=[1-1/(1+2)2:2][1-1/(1+3)3:2]..........[1-1/(1+20006)2006:2]=>(1-1/3)(1-1/6).........(1-2/2006x2007)=>A=2/3.5/6.9/10.......2006.2007-2/2006.2007=>4/6.10/2.18/20.....2006.2007-2/2006.2007(1) . Ta lai co : 2007.2006-2=[2006.(2008-1)+2006]-2008=2006.(2008-1
+1)-2008=2006.2008-2008=2008.(2006-1)=2005.2008(2) . Tu (1) va (2) ta co : A = 4/6.10/12.18/20.....2005.2008/2006.2007=>A=4.1/2.3.2.5/4.3.6.3/4.5........2005.2008/2007.2006=>A=(1.2.3....2005)(4.5.6....2008)/(2.3.4.....2006)(3.4.5...2007)=2008/2006.3
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A= \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2005\cdot2006}\)
A= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2005}-\dfrac{1}{2006}\)
A= \(1-\dfrac{1}{2006}\)
A= \(\dfrac{2005}{2006}\)
Vậy A= \(\dfrac{2005}{2006}\)
- Đặt \(A=1-\frac{1}{2^2}-\frac{1}{3^2}-...-\frac{1}{2006^2}\)
- Ta có: \(1=1\)
\(\frac{1}{2^2}>\frac{1}{2.3}\)
\(\frac{1}{3^2}>\frac{1}{3.4}\)
\(................\)
\(\frac{1}{2006^2}>\frac{1}{2006.2007}\)
\(\Rightarrow A>1-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{2006.2007}\)
\(\Leftrightarrow A>1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)-...-\left(\frac{1}{2006}-\frac{1}{2007}\right)\)
\(\Leftrightarrow A>1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-...-\frac{1}{2006}+\frac{1}{2007}\)
\(\Leftrightarrow A>1+\frac{1}{2007}=\frac{2008}{2007}\)mà \(\frac{2008}{2007}>1>\frac{1}{2006}\)
\(\Rightarrow A>\frac{1}{2006} \left(ĐPCM\right)\)
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