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\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1003}\right)\)
\(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2016}\)
Đặt A=1-1/2+1/3-1/4+.......+1/2005-1/2006
=>A= (1+1/3+1/5+...+1/2005)-(1/2+1/4+1/6+.....+1/2006)
=>A=(1+1/2+1/3+...+1/2005)-2.(1/2+1/4+1/6+...+1/2006)
=>A=(1+1/2+1/3+....+1/2005)-(1+1/2+1/3+...+1/1003)
=>A=1/1004+1/1005+.....+1/2006
Vậy A=1/1004+1/1005+.....+1/2006 ( Điều phải chứng minh )
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:
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2005.2008}{2006.2007}\)
\(=\frac{(1.2.3...2005)(4.5.6...2008)}{(2.3.4...2006)(3.4.5...2007)}=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)
Ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow\frac{1}{3}.B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)
\(\Rightarrow B-\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(\frac{2}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(B=\left(\frac{1}{3}-\frac{1}{3^{2006}}\right):\frac{2}{3}\)
\(B=\frac{1}{3}:\frac{2}{3}-\frac{1}{3^{2006}}:\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)