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Áp dụng công thức:
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\) thì được
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2009}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2009.2010}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2009.2010}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{1004}{1005}\)
thôi, làm luôn nè
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2009}\)
\(=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2009\right).2009:2}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2009.2010}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+2.\left(\frac{1}{4}-\frac{1}{5}\right)+...+2.\left(\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
\(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}.\) Nhân với 2 cả hai vế:
được: \(2M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\) Suy ra : \(M=2M-M=1-\frac{1}{2^{100}}\)
CHÚC BẠN HỌC GIỎI
\(\frac{\left(\frac{1}{2}\right)^2.2018-\left(\frac{1}{4}\right)^2.2017}{\frac{1}{4096}.\frac{1}{3}+2^{13}}\)
=
\(\frac{\left(\frac{1}{2}\right)^2.2018-\left(\frac{1}{4}\right)^6.2017}{\frac{1}{4096}.\frac{1}{3}+2^{13}}\)\(\Leftrightarrow\frac{\left(\frac{1}{4}\right).2018-\left(\frac{1}{4096}\right).2017}{\frac{1}{4096}.\frac{1}{3}+2^{13}}\)
Lược bỏ các số giống nhau đi ta được :
\(\frac{\left(\frac{1}{4}\right).2018.2017}{\frac{1}{3}+2^{13}}\Leftrightarrow\frac{\left(\frac{1}{4}\right).2018.2017}{\frac{1}{3}.8192}\Leftrightarrow\frac{\frac{1}{4}.4070306}{\frac{8192}{3}}\)
\(=\frac{1017576,5}{\frac{8192}{3}}\)
B=2013.(1+
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)
B=2013(\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
B=2013.2(\(1\frac{1}{2013}=2013.2.\frac{2012}{2013}=4024\)
\(2M=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2M-M=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{10}}\)
\(=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)
2M=1+1/2+1/2^2+...+1/2^9
M=2M-M= 1/2-1/2^10(triệt tiêu mấy cái giống nhau nha)
M=(2^9-1)/2^10
Nè :33