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27 tháng 3 2018
Ta có:
\(\frac{a_1}{a_2}=\frac{a_2}{a_3};\frac{a_2}{a_3}=\frac{a_3}{a_4};...;\frac{a_{2015}}{a_{2016}}=\frac{a_{2016}}{a_{2017}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{2016}}{a_{2017}}=k\)
\(\Rightarrow\frac{a_1^{2016}}{a_2^{2016}}=\frac{a_2^{2016}}{a_3^{2016}}=...=\frac{a_{2016}^{2016}}{a_{2017}^{2016}}=\frac{a_1^{2016}+a_2^{2016}+...+a_{2016}^{2016}}{a_2^{2016}+a_3^{2016}+...+a_{2017}^{2016}}=k^{2016}\left(1\right)\)
Ta lại có:
\(k^{2016}=\frac{a_1}{a_2}.\frac{a_2}{a_3}...\frac{a_{2016}}{a_{2017}}=\frac{a_1}{a_{2017}}\left(2\right)\)
Từ (1) và (2) \(\frac{a_1^{2016}+a_2^{2016}+...+a_{2016}^{2016}}{a_2^{2016}+a_3^{2016}+...+a_{2017}^{2016}}=\frac{a_1}{a_{2017}}\)
14 tháng 11 2018
\(P=\frac{3}{1!\left(1+2\right)+3!}+\frac{4}{2!\left(1+3\right)+4!}+...+\frac{2017}{2015!\left(1+2016\right)+2017!}\)
\(P=\frac{3}{3\left(1!+2!\right)}+\frac{4}{4\left(2!+3!\right)}+...+\frac{2017}{2017\left(2015!+2016!\right)}\)
\(P=\frac{1}{1!+2!}+\frac{1}{2!+3!}+...+\frac{1}{2015!+2016!}\)
Ta có \(a!>\sqrt{a}\)\(\left(a\inℕ;a>1\right)\) do đó :
\(P>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2015}+\sqrt{2016}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\)
\(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}+\sqrt{2015}\right)\left(\sqrt{2016}-\sqrt{2015}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2016}\)
\(-\sqrt{2015}=\sqrt{2016}-1=\frac{1}{2}+\left(\sqrt{2016}-\frac{3}{2}\right)=\frac{1}{2}+\left(\sqrt{2016}-\sqrt{\frac{9}{4}}\right)>\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Chúc bạn học tốt ~
PS : tự nghĩ bừa thui nhé :))
AA
8 tháng 3 2018
Ta có A= 1/2015 + 2/2016 + 3/2017 + ... +2016/4030- 2016
A= 2015-2014/2015 + 2016-2014/2016 +...+4030-2014/4030-2016
A= 2015/2015-2014/2015+ 2016/2016-2014/2016 + ..... +4030/4030-2014/4030 -2016
A= 1-2014/2015 + 1-2014/2016 +....+1-2014/4030 -2016
A= (1+1+1+1+........+1) -(2014/2015+2014/2016+......+2014/4030) -2016
A=2016 - 2014.(1/2015+1/2016+....+1/4030) -2016
A= (2016 - 2016 ) - 2014. ( 1/2015+1/2016+.....+1/4030)
A=-2014.(1/2015+1/2016+....+1/4030)
mà B = 1/2015+1/2016+....+1/4030
nên A : B = -2014
HT
1
6 tháng 9 2017
\(M=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}\)
\(M=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)\(M=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2008}\right)\)
\(M=\dfrac{1}{2009}+\dfrac{1}{2010}+...+\dfrac{1}{2016}+\dfrac{1}{2017}=N\)
Vậy \(\left(M-N\right)^{2017}=0\)
NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)