Violympic toán 6

TL

Tính:

a) \(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)b) \(B=\dfrac{1-3}{1\cdot3}+\dfrac{2-4}{2\cdot4}+\dfrac{3-5}{3\cdot5}+\dfrac{4-6}{4\cdot6}+...+\dfrac{2011-2013}{2011\cdot2013}+\dfrac{2012-2014}{2012\cdot2014}+\dfrac{2013-2015}{2013\cdot2015}\)Giúp mình với!

LB
20 tháng 4 2018 lúc 10:17

\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)

\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)

\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)

\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)

\(2A=2+3+4+5+6+...+2012+2013+2014\)

\(2A=\dfrac{\left(2+2014\right).2013}{2}\)

\(A=\dfrac{2016.2013}{4}=504.2013\)

Bình luận (2)
LB
20 tháng 4 2018 lúc 10:40

\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)

\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)

\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)

\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)

\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)

\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)

\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)

\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)

Bình luận (1)

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