1/3 + 1/3^2 +1/3^3+...+1/3^2014<1/2
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MC
0
VN
3
27 tháng 12 2017
A = 1 + 2014^1 + 2014^2 + 2014^3 + ... + 2014^2014 + 2014^2015
2014A = 2014^1 + 2014^2 + 2014^3 + 2014^4 + ... 2014^2015 + 2014^2016
2014A - A = ( 2014^1 + 2014^2 + 2014^3 + 2014^4 + .... + 2014^2015 + 2014^2016 ) - ( 1 + 2014^1 + 2014^2 + 2014^3 + ... + 2014^2014 + 2014^2015 )
2013A = 2014^2016 - 1
A = 2014^2016 - 1 / 2013
B = 3 - 3^2 + 3^3 + 3^4 + ... + 3^100 ( đề hơi vui )
3B = 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101
3B - B = ( 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101 ) - ( 3 - 3^2 + 3^3 + 3^4 + ... + 3^100 )
2B = ( 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101 ) - 3 + 3^2 - 3^3 - 3^4 - ... - 3^100
2B = 3^2 - 3^3 + 3^101 - 3 + 3^2 - 3^3
2B = 9 - 27 + 3^101 - 3 + 9 - 27
2B = -18 + 3^101 - 3 + ( -18 )
2B = -39 + 3^101
B = -39 + 3^101 / 2
27 tháng 12 2017
A = 1 + 2014 + 20142 + 20143 + ... + 20142014 + 20142015
2014A = 2014 + 20142 + 20143 + 20144 + ... + 20142015 + 20142016
2014A - A = ( 2014 + 20142 + 20143 + 20144 + ... + 20142015 + 20142016 ) - ( 1 + 2014 + 20142 + 20143 + ... + 20142014 + 20142015 )
2013A = 20142016 - 1
A \(=\frac{2014^{2016}-1}{2013}\)
TC
0
12 tháng 3 2019
\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
TN
1
31 tháng 10 2016
\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)
\(P=\frac{\left(1+2\right).2:2-1}{\left(1+2\right).2:2}.\frac{\left(1+3\right).3:2-1}{\left(1+3\right).3:2}.\frac{\left(1+4\right).4:2-1}{\left(1+4\right).4:2}...\frac{\left(1+2014\right).2014:2-1}{\left(1+2014\right).2014:2}\)
\(P=\frac{2}{2.3:2}.\frac{5}{3.4:2}.\frac{9}{4.5:2}...\frac{2029104}{2014.2015:2}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2013.2016}{2014.2015}\)
\(P=\frac{1.2.3...2013}{2.3.4...2014}.\frac{4.5.6...2016}{3.4.5...2015}\)
\(P=\frac{1}{2014}.\frac{2016}{3}=\frac{1}{2014}.672=\frac{336}{1007}\)
NT
0
\(\frac{A}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2015}}.\)
\(\frac{2A}{3}=A-\frac{A}{3}=\frac{1}{3}-\frac{1}{3^{2015}}=\frac{3^{2014}-1}{3^{2015}}\)
\(A=\frac{3\left(3^{2014}-1\right)}{2.3^{2015}}=\frac{3^{2015}-3}{2.3^{2015}}=\frac{1}{2}-\frac{1}{2.3^{2014}}< \frac{1}{2}\)