Tính A = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{2013}\left(1+2+...+2013\right)\)
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AH
Akai Haruma
Giáo viên
24 tháng 8
Lời giải:
** Sửa đề:
$A=\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+\frac{1}{4}(1+2+3+4)+....+\frac{1}{2013}(1+2+3+...+2013)$
$A=\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+....+\frac{1}{2013}.\frac{2013.2014}{2}$
$=\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2014}{2}$
$=\frac{3+4+5+...+2014}{2}$
$=\frac{1+2+3+4+5+...+2014}{2}-\frac{3}{2}$
$=\frac{2014.2015:2}{2}-\frac{3}{2}$
$=1014551$
17 tháng 9 2017
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)( có 2013 thừa số )
\(A=\left(-\frac{3}{2^2}\right).\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right).....\left(-\frac{\text{4056196}}{2014^2}\right)\)
\(-A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{4056196}{2014^2}=\frac{1.3.2.4.3.5....2013.2015}{2.2.3.3.4.4.....2014.2014}\)
\(-A=\frac{\left(1.2.3...2013\right).\left(3.4.5.6...2015\right)}{\left(2.3.4.5....2014\right).\left(2.3.4.5...2014\right)}=\frac{1.2015}{2.2014}=\frac{2015}{4028}\)
\(A=-\frac{2015}{4028}\)
Vậy.....
Z
17 tháng 9 2017
-A=(\(1-\frac{1}{2^2}\)) . (\(1-\frac{1}{3^2}\))......(\(1-\frac{1}{2014^2}\))
-A= \(\frac{3}{4}\). \(\frac{8}{9}\). ...... \(\frac{4056195}{4056196}\)
-A= \(\frac{1.3.2.4.......2013.2015}{2.2.3.3.......2.14.2014}\)
-A= \(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\)
-A= \(\frac{2015}{2014.2}\)
-A=\(\frac{2015}{4028}\)
28 tháng 7 2019
Ta có:\(\left(x-1\right)\left(x+1\right)=x\left(x-1\right)+x-1^2=x^2-x+x-1=x^2-1\)
Áp dụng:\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{2014^2-1}{2014\cdot2014}\)
\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot...\cdot\frac{2013\cdot2015}{2014^2}\)
\(=\frac{1}{2}\cdot\frac{2015}{2014}=\frac{2015}{4028}\)
LH
13 tháng 8 2016
Ta áp dụng công thức: \(a-b=\left[-\left(b-a\right)\right]\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=-\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2012}\right)\left(1-\frac{1}{2013}\right)\right]\)
\(=-\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2011}{2012}.\frac{2012}{2013}\right)\)
\(=-\frac{1.2.3...2011.2012}{2.3.4....2012.2013}\)
\(=-\frac{1}{2013}\)
DS
13 tháng 8 2016
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2012}{2013}\)
Liệt tử thừa với mẫu thừa:
\(=\frac{1}{2013}\)
Chúc em học tốt^^
1 tháng 11 2016
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-2011}{2012}.\frac{-2012}{2013}\)
\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-2011\right).\left(-2012\right)}{2.3.4....2013}\)
\(=\frac{1.2.3...2011.2012}{2.3.4.5...2013}\) ( vì các số hạng ở trên tử chẵn )
\(=\frac{1}{2013}\)
theo công thức \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
=>\(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{2013}.\frac{2013.2014}{2}\)
\(=>A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2014}{2}=>A=\frac{1}{2}\left(1+2+3+..+2014\right)-\frac{1}{2}\)
\(=>A=\frac{1}{2}.\frac{2014.2015}{2}-\frac{1}{2}=1014552\)