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NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
14 tháng 8 2016
a)\(=\frac{2017}{2016}.\frac{3}{4}-\frac{1}{2016}.\frac{3}{4}\)
\(=\frac{3}{4}\left(\frac{2017}{2016}-\frac{1}{2016}\right)\)
\(=\frac{3}{4}.1\)
\(=\frac{3}{4}\)
b)\(=\frac{2015}{2016}\left(\frac{1}{2}+\frac{1}{3}-\frac{5}{6}\right)\)
\(=\frac{2015}{2016}.0\)
\(=0\)
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
HT
31 tháng 10 2015
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=....=\frac{a_{2015}}{a_{2016}}=\frac{a_1+a_2+...+a_{2015}}{a_2+a_3+...+a_{2016}}\)
=> \(\left(\frac{a_1+a_2+....+a_{2015}}{a_2+a_3+....+a_{2016}}\right)^{2015}=\frac{a_1.a_2.....a_{2015}}{a_2.a_3......a_{2016}}=\frac{a_1}{a_{2016}}\)
=> \(\left(\frac{a_1+a_2+....+a_{2015}}{a_2+a_3+....+a_{2016}}\right)^{2015}=\frac{a_1}{a_{2016}}\)(Đpcm)
NV
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
A= 3+3^2+3^3+.....+3^2015+3^2016
2A=3^2+3^4+........+3^2016 +2^2017
2A-A= (3^2-3^2) + ( 3^3-3^3)+..........+(3^2015-3^2015)+(3^2016-3^2016)+(3^2017 -3)
A= 3 ^2017 - 3
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