\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2011}\left(1+2+3+...+2011\right)\)
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28 tháng 12 2016
Tổng các số tự nhiên từ 1 đến n là \(\frac{n\left(n+1\right)}{2}\)
Do đó \(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2011}.\frac{2011.2012}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2012}{2}\)
\(=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\right)-\frac{1}{2}\)
\(=\frac{1+2+3+...+2012}{2}-\frac{1}{2}\)
\(=\frac{\frac{2012.2013}{2}}{2}-\frac{1}{2}\)
\(=1012538,5\)
Vậy ....
6 tháng 5 2017
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
Vậy x=1
6 tháng 5 2017
\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)
\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)
\(\frac{1}{2011}.x=\frac{2}{4022}\)
\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)
Ai thấy đún thì ủng hộ mink nha !!!
Thanks you very much !!
Chúc các bạn luôn học giỏi !!!
17 tháng 5 2019
C=(1+2/3).(1+2/5).(1+2/7)......(1+2/2009).(1+2/2011)
C=5/3.7/5.9/7......2011/2009.2013/2011
C=5.7.9.....2013/3.5.7.....2009.2011
C=2013/3
N
20 tháng 1 2019
\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
24 tháng 2 2018
=\(\frac{2}{1+2}.\frac{2+3}{1+2+3}.\frac{2+3+4}{1+2+3+4}...\frac{2+3+4+...+2011}{1+2+3+....+2011}\)
=\(\frac{2}{\frac{\left(2+1\right).2}{2}}.\frac{\left(2+3\right).2}{\frac{2}{\frac{\left(3+1\right).3}{2}}}....\frac{\left(2+2011\right)\left(2011-1\right)}{\frac{2}{\frac{\left(2011+1\right)2011}{2}}}\)
=\(\frac{4}{\left(2+1\right).2}\frac{\left(2+3\right).2}{\left(3+1\right).3}....\frac{(2+2011)\left(2011-1\right)}{\left(2011+1\right)2011}\)
=\(\frac{\left(1.4\right)\left(5.2\right)....\left(2013.2010\right)}{\left(3.2\right).\left(4.3\right)....\left(2012.2011\right)}\)
=\(\frac{\left(1.2.3...2010\right)\left(4.5.6...2013\right)}{\left(2.3.4...2011\right)\left(3.4.5....2012\right)}\)
=\(\frac{1}{2011}.\frac{2013}{3}\)=\(\frac{671}{2011}\)
Mk nghĩ vậy. Chắc là đúng đấy
18 tháng 4 2018
Suy ra : A = ( 1 - 1 / 2010 ) . ( 1 - 2 / 2010 ) .... 0 . ( 1 - 2011 / 2010 ) = 0
Suy ra A = 0
18 tháng 4 2018
A = 1. ( 1/2010 + 2/2010 ) - ( 3/2010 + 4/2010 ) - ... - ( 2010/2010 + 2011/2010 )
= 1/2010 - 2011/2010
= -2010/2010
\(\frac{1+2+...+n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n+1}{2}\)
\(\Rightarrow A=1+\frac{1}{2}\left(3+4+...+2012\right)\)
\(=1+\frac{1}{2}\left(1+2+...+2012-3\right)\)
\(=1+\frac{1}{2}\left(1+2+...+2012\right)-\frac{3}{2}\)
\(=\frac{1}{2}.\frac{2012.2013}{2}-\frac{1}{2}=503.2013-\frac{1}{2}=...\)
1)503x2013
hay
2)503,2013
hả bạn nó là nhân hay phẩy