Biểu thức A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+......+\frac{1}{1+2+3+4+...2014}\) có giá trị bằng
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15 tháng 8 2016
Ta có : \(1+2=\frac{2.3}{2}\) , \(1+2+3=\frac{3.4}{2}\) ,
\(1+2+3+4=\frac{4.5}{2}\) , ......... , \(1+2+3+4+....+2014=\frac{2014.2015}{2}\)
Suy ra : \(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2014.2015}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(2\left(\frac{1}{2}-\frac{1}{2015}\right)\)
15 tháng 8 2016
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}\)
\(A=\frac{2013}{2015}\)
HB
21 tháng 1 2016
A=2/6+2/12+....+2/4054182
A=2/2.3+2/3.4+...+2/2013.2014
A= (1-2/2014) : 2=503/1007
25 tháng 6 2016
A =\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+...+\(\frac{1}{1+2+3+4...+2014}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\)
\(\Rightarrow2A=2\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\right)\)
\(\Rightarrow2A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4058210}\)
\(\Rightarrow2A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{2013}{4030}\)
\(\Rightarrow A=\frac{2013}{8060}\)
1 tháng 7 2016
ngài Kiệt ღ ๖ۣۜLý๖ۣۜ đúng là không ái sánh bằng sự gian xảo này
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}=\frac{2013}{2015}\)