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A = 1.3 + 2.4 + 3.5 + 4.6 + 5.7 + .. + 2013.2015 = [1.3 + 3.5+..+2013.2015] + [2.4 + 4.6 + .. + 2012.2014] = X + Y
X = 1.3 + 3.5 + 5.7 + .. + 2013.2015
X.6 = 1.3.(5 - (-1)) + 3.5.(7 - 1) + 5.7.(9-3) + 7.9.(11-5) + .. + 2011.2013.(2015-2009) + 2013.2015.(2017-2011)
= -(-1).1.3 + 1.3.5 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + .... = 1.3 + 2013.2015.2017
=> X = 1/6*(3 + 2013.2015.2017) = ...
tương tự
Y = 2.4 + 4.6 + .. + 2012.2014
Y.6 = 2.4.6 + 4.6.(8-2) +... + 2012.2014.(2016-2010) = 2012.2014.2016
=> Y = 2012.2014.2016/6 = ...
=> A = X + Y = 2725086001
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2013.2015}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)
Vậy A=1007/2015
\(2A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(2A=1-\frac{1}{2015}\)
\(A=\frac{2014}{2015}:2\)
\(A=\frac{1007}{2015}\)
A = 1.3 + 2.4 + 3.5 + 4.6 + 5.7 + .. + 2013.2015 = [1.3 + 3.5+..+2013.2015] + [2.4 + 4.6 + .. + 2012.2014] = X + Y
X = 1.3 + 3.5 + 5.7 + .. + 2013.2015
X.6 = 1.3.﴾5 ‐ ﴾‐1﴿﴿ + 3.5.﴾7 ‐ 1﴿ + 5.7.﴾9‐3﴿ + 7.9.﴾11‐5﴿ + .. + 2011.2013.﴾2015‐2009﴿ + 2013.2015.﴾2017‐2011﴿
= ‐﴾‐1﴿.1.3 + 1.3.5 + 3.5.7 ‐ 1.3.5 + 5.7.9 ‐ 3.5.7 + .... = 1.3 + 2013.2015.2017
=> X = 1/6*﴾3 + 2013.2015.2017﴿ = 1363557553
tương tự Y = 2.4 + 4.6 + .. + 2012.2014
Y.6 = 2.4.6 + 4.6.﴾8‐2﴿ +... + 2012.2014.﴾2016‐2010﴿ = 2012.2014.2016
=> Y = 2012.2014.2016/6 = 1361528448
=> A = X + Y = 2725086001
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