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6 tháng 9 2016
Xét mẫu số của F :
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2016\cdot2017}{2}}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{2017}\right)=2-\frac{2}{2017}=\frac{4032}{2017}\)
Suy ra : \(F=\frac{2.2016}{\frac{4032}{2017}}=\frac{2.2016.2017}{4032}=2017\)
DP
14 tháng 8 2017
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
8 tháng 1 2016
A = 1.2+2.3+...+2016.2017
3A=1.2.3 + 2.3.(4-1) + .. + 2016.2017.(2018-2015)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 2016.2017.2018 - 2015.2016.2017
3A = 2016.2017.2018
A = 2016.2017.2018 : 3
A = 2735245632
8 tháng 1 2016
3A=1*2*3+2*3*(4-1)+.........+2016*2017.(2018-2015)
3A=1.2.3-1.2.3+2.3.4-2.3.4+.........+2016.2017.3
3A=2016.2017.2018
KẾT QUẢ TỰ TÍNH
NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
GT
1
28 tháng 10 2016
Đặt \(A=1+3+3^2+3^3+....+3^{2015}-3^{2016}\)
\(B=1+3+3^2+3^3+....+3^{2015}\)
Ta có:
\(B=1+3+3^2+...+3^{2015}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{2016}\)
\(\Rightarrow3B-B=\left(3+3^2+3^3+...+3^{2016}\right)-\left(1+3+3^2+...+3^{2015}\right)\)
\(\Rightarrow2B=3^{2016}-1\)
\(\Rightarrow B=\frac{3^{2016}-1}{2}\)
\(\Rightarrow A=\frac{3^{2016}-1}{2}-3^{2016}\)
3A=1/3^2-1/3^3+...-1/3^2017
3A+A=1-1/3+...-1/3^2015+1/3-1/3^2+...-1/3^2016
4A=1-1/3^2016
4A=3^2016-1/3^2016
A=3^2016-1/4.3^2016