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b) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
A= 1/3 + 1/3^2 + ... + 1/3^8
3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)
3A=1+ 1/3 + 1/3^2+ ... +1/3^7
=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)
=> 2A= 1 - 1/ 3^8
2A= 6560/6561
A= 6560/6561 : 2
A= 3280/6561
Ta có :
\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=2009\)
\(A=-B\)
\(B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{23.25}+\dfrac{2}{25.27}+\dfrac{1}{27}\)
\(B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{23}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}\)
\(B=1\)
A=-1
\(A=-\dfrac{2}{1.3}-\dfrac{2}{3.5}-......-\dfrac{2}{25.27}-\dfrac{1}{27}\)
\(\Leftrightarrow A=-\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+.....+\dfrac{1}{27}\right)\)
\(\Leftrightarrow A=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}\right)\)
\(\Leftrightarrow A=-1\)
Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{2005}}\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2A=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)
\(2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2004}}\)
\(2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2004}}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{2005}}\right)\)
\(A=2-\dfrac{1}{2^{2005}}\)
Giải:
Ta có: A = \(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^{2005}}.\)
= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...=\dfrac{1}{2^{2005}}.\)
2A = \(2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right).\)
= \(1+2+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}.\)
2A -A = \(\left(1+2+\dfrac{1}{2}+...+\dfrac{1}{2^{2004}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right).\)
= 2 - \(\dfrac{1}{2^{2005}}.\)
Vậy \(A=2-\dfrac{1}{2^{2005}}.\)
CHÚC BN HỌC TỐT!!! ^-^
Đừng quên bình luận nếu bài mik sai nhé!!!
