AH
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K
Khách
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TV
1
21 tháng 9 2016
Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)
A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)
2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)
2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)) \(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))
A=\(1+\frac{1}{2^{10}}\)
A= \(\frac{1025}{1024}\)
LT
0
LM
7 tháng 9 2016
\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-....-\frac{1}{6.4}-\frac{1}{4.2}\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100.98}+\frac{1}{98.96}+....+\frac{1}{6.4}+\frac{1}{4.2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{98}+\frac{1}{98}-\frac{1}{96}+.....+\frac{1}{6}-\frac{1}{4}+\frac{1}{4}-\frac{1}{2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)\Rightarrow A=\frac{1}{100}-\frac{1}{100}+\frac{1}{2}\Rightarrow A=\frac{1}{2}\)
7 tháng 9 2016
\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-...-\frac{1}{6.4}-\frac{1}{4.2}\)
\(A=\frac{1}{100}-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{96.98}+\frac{1}{98.100}\right)\)
\(A=\frac{1}{100}-\frac{1}{2.2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\frac{49}{50}\)
\(A=\frac{2}{200}-\frac{49}{200}=-\frac{47}{200}\)
7 tháng 4 2018
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+......+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+........+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+......+\frac{1}{2013}\)
\(=P\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0\)