Tính nhanh S=3/1.3+3/3.5+..+3/2013.2015
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TZ
1
12 tháng 11 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) = ...
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
11 tháng 12 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
lâu ko làm nên sai hay đúng thì mình ko biết nữa
VM
1
HN
16 tháng 5 2021
\(B=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2013\times2015}\\
2B=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2013\times2015}\\
2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\\2B=1-\frac{1}{2015}=\frac{2014}{2015}\\
\Rightarrow B=\frac{2014}{2015}
\div2=\frac{1007}{2015}\)
TA
2
HP
12 tháng 5 2016
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2013.2015}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)
Vậy A=1007/2015
TN
12 tháng 5 2016
\(2A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(2A=1-\frac{1}{2015}\)
\(A=\frac{2014}{2015}:2\)
\(A=\frac{1007}{2015}\)
12 tháng 5 2016
1/1.3+1/3.5+...+1/2013.2015
=1/2.(1/1-1/3+1/3-1/5+...+1/2013-1/2015)
=1/2.(1/1-1/2015)
=1/2.2014/2015
=1007/2015
12 tháng 5 2016
A=1/1.3+1/3.5+1/5.7+...+1/2013.2015
2A=2.(1/1.3+1/3.5+1/5.7+...+1/2013.2015)
=2/1.3+2/3.5+2/5.7+...+2/2013.2015
=1-1/3+1/5-1/7+1/7-1/9+...+1/2013-1/2015
=1-1/2015
=2014/2015
=>2A=2014/2015=>A=1007/2015
NM
1
KB
21 tháng 5 2017
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}\)
=\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{2015}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2016}\right)\)
=\(\frac{3}{4}-\left(\frac{1}{4030}+\frac{1}{4032}\right)\) < \(\frac{3}{4}\)
=> đpcm
\(S=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{2013.2015}\)
\(S=3.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)
\(S=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)
\(S=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(S=\frac{3}{2}.\left(1-\frac{1}{2015}\right)\)
\(S=\frac{3}{2}.\frac{2014}{2015}\)
\(S=\frac{2021}{2015}\)