Tính giá trị biểu thức
A =\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+......+\frac{1}{2014^2}\)
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18 tháng 8 2020
\(2014:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1\frac{2}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{1\frac{1}{6}+0,875-0,7}{\frac{1}{3}+0,25-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}+\frac{2}{8}-\frac{2}{10}}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7}{2}\right)=2014\)
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
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}\)
\(A=\frac{2013}{2015}\)
TX
1
13 tháng 8 2016
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}=\frac{2013}{2015}\)
10 tháng 8 2017
Mỗi biểu thức trong dấu căn có dạng:
\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\) ( Với \(k\ge2\))
Ta có:
\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}=\frac{k^4+2k^3+k^2+k^2+2k+1+k^2}{k^2\left(k+1\right)^2}\)
\(=\frac{k^4+2k^2\left(k+1\right)+\left(k+1\right)^2}{k^2\left(k+1\right)^2}=\frac{\left(k^2+k+1\right)^2}{\left(k\left(k+1\right)\right)^2}\)
\(\Rightarrow\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{k^2+k+1}{k^2+k}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2013}-\frac{1}{2014}=2014-\frac{1}{2014}\)
B
22 tháng 9 2017
Mỗi biểu thức trong dấu căn có dạng:
1+1k2 +1(k+1)2 ( Với k≥2)
Ta có:
1+1k2 +1(k+1)2 =k2(k+1)2+(k+1)2+k2k2(k+1)2 =k4+2k3+k2+k2+2k+1+k2k2(k+1)2
=k4+2k2(k+1)+(k+1)2k2(k+1)2 =(k2+k+1)2(k(k+1))2
⇒√1+1k2 +1(k+1)2 =k2+k+1k2+k =1+1k(k+1) =1+1k −1k+1
⇒S=1+1−12 +1+12 −13 +1+13 −14 +...+1+12013 −12014 =2014−12014
\(\approx0,4\)