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A = 2013 + \(\frac{2013}{1+2}\)+\(\frac{2013}{1+2+3}\)+\(\frac{2013}{1+2+3+4}\)+ ..........+\(\frac{2013}{1+2+3+....+2012}\)
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Cô Hoàng Huyền
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8 tháng 5 2017
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
10 tháng 11 2016
\(D=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}\)
\(=\frac{2013}{2014}\)
19 tháng 3 2018
B=2013.(1+
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)
B=2013(\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
B=2013.2(\(1\frac{1}{2013}=2013.2.\frac{2012}{2013}=4024\)
4 tháng 5 2016
=> B=2013. (1+\(\frac{1}{1+2}\) +\(\frac{1}{1+2+3}\) +...+ \(\frac{1}{1+2+3+...+2012}\))
=>B= 2013.(\(\frac{2}{2}\) + \(\frac{2}{2.3}\) +\(\frac{2}{3.4}\) +...+\(\frac{2}{2012.2013}\))
=>B= 2013.2.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +\(\frac{1}{3.4}\) +...+\(\frac{1}{2012.2013}\))
=>B=4026. (1-\(\frac{1}{2}\) +\(\frac{1}{2}\) -\(\frac{1}{3}\) + ...+\(\frac{1}{2012}\) - \(\frac{1}{2013}\))
=>B=4026.(1-\(\frac{1}{2013}\))
=>B=4026.\(\frac{2012}{2013}\) => B=2.2012=4024 Vậy B=4024