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\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+...+\frac{2+3+...+20}{1+2+3+...+20}\)
\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)
\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)
\(A=\left(1+1+....+1\right)\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{210}\right)\)
\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)
\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)
\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\)
\(A=19-2\cdot\frac{19}{42}=19-\frac{19}{21}=\frac{380}{21}\)
Vậy A= \(\frac{380}{21}\)
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2005}\right)\left(1-\frac{1}{2006}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2004}{2005}\cdot\frac{2005}{2006}\)
\(B=\frac{1\cdot2\cdot...\cdot2004\cdot2005}{2\cdot3\cdot...\cdot2005\cdot2006}\)
\(B=\frac{1}{2006}\)
Vậy \(B=\frac{1}{2006}\)
Lời giải:
Xét công thức tổng quát:
$1+2+3+...+n=\frac{n(n+1)}{2}$
$\Rightarrow 1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$
Thay $n=2,3,...,2006$ ta thu được:
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2005.2008}{2006.2007}\)
\(=\frac{(1.2.3...2005)(4.5.6...2008)}{(2.3.4...2006)(3.4.5...2007)}=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)
Lời giải:
Xét công thức tổng quát:
$1+2+3+...+n=\frac{n(n+1)}{2}$
$\Rightarrow 1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$
Thay $n=2,3,...,2006$ ta thu được:
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2005.2008}{2006.2007}\)
\(=\frac{(1.2.3...2005)(4.5.6...2008)}{(2.3.4...2006)(3.4.5...2007)}=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)
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= \(\dfrac{2005}{2006}\)
Nếu \(n>0\Rightarrow\left(n-1\right)n\left(n+1\right)=n^3-n< n^3.\)
\(\Rightarrow VT< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2005.2006.2007}\)
\(\Rightarrow2.VT< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2005.2006.2007}\)
\(\Rightarrow2.VT< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2007-2005}{2005.2006.2007}\)
\(\Rightarrow2VT< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\)
\(\Rightarrow2.VT< \frac{1}{2}-\frac{1}{2006.2007}\Rightarrow VT< \frac{1}{4}-\frac{1}{2.2006.2007}< \frac{1}{4}\)
Đặt: \(L_2=\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
\(L_2=1+\left(\dfrac{2006}{2}+1\right)+\left(\dfrac{2005}{3}+1\right)+...+\left(\dfrac{2}{2006}+1\right)+\left(\dfrac{1}{2007}+1\right)\)
\(L_2=\dfrac{2008}{2008}+\dfrac{2008}{2}+\dfrac{2008}{3}+...+\dfrac{2008}{2006}+\dfrac{2008}{2007}\)
\(L_2=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(\dfrac{L_1}{L_2}=\dfrac{1}{2008}\)
A= \(\left(1-\frac{1}{1+2}\right)\)\(\left(1-\frac{1}{1+2+3}\right)\) \(\left(1-\frac{1}{1+2+3+4}\right)\) .....\(\left(1-\frac{1}{1+2+3+...+2005+2006}\right)\)
A = \(\left(1-\frac{1}{3}\right)\) \(\left(1-\frac{1}{6}\right)\) \(\left(1-\frac{1}{10}\right)\) .... \(\left(1-\frac{1}{2013021}\right)\)
= \(\frac{2}{3}\) . \(\frac{5}{6}\) . \(\frac{9}{10}\) .....\(\frac{2013020}{2013021}\)
= \(\frac{4}{6}\).\(\frac{10}{12}\).\(\frac{18}{20}\)....\(\frac{4026040}{4026042}\)
= \(\frac{1.4}{2.3}\).\(\frac{2.5}{3.4}\).\(\frac{3.6}{4.5}\).\(\frac{2005.2008}{2006.2007}\)
= \(\frac{1.2.3.4...2005}{2.3.4.5...2006}\).\(\frac{4.5.6...2008}{3.4.5...2007}\)
= \(\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)
Đề bài là A = gì thế bạn?