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ta có
a1+(a2+a3+a4)+... +(a11+a12+a13)+a14+(a15+a16+a17)+(a18+a19+a20)<0
a1>0; a2+a3+a4>0;...;a11+a12+a13>0;a15+a16+a17>0;a18+a19+a20>0; a14<0
Ta có:
(a1+a2+a3)+...+(a10+a11+a12)+(a13+a14)+(a15+a16+a17)+(a18+a19+a20)<0
=>(a13+a14)<0
có a12+a13+a14>0=>a12>0
Từ các cmt suy ra a1>0; a12>0; a14<0
=>a1. a14+a12.a12<a1.a12(đpcm)
# HOK TỐT #
ta có
a1+(a2+a3+a4)+... +(a11+a12+a13)+a14+(a15+a16+a17)+(a18+a19+a20)<0
a1>0; a2+a3+a4>0;...;a11+a12+a13>0;a15+a16+a17>0;a18+a19+a20>0; a14<0
Ta có:
(a1+a2+a3)+...+(a10+a11+a12)+(a13+a14)+(a15+a16+a17)+(a18+a19+a20)<0
=>(a13+a14)<0
có a12+a13+a14>0=>a12>0
Từ các cmt suy ra a1>0; a12>0; a14<0
=>a1. a14+a12.a12<a1.a12
Ta thấy : \(a_1+a_2+a_3+.....+a_{2015}+a_1=1008.1=1008\)
Mà \(a_1+a_2+a_3+......+a_{2015}=0\)
\(\Rightarrow a_1+\left(a_1+a_2+a_3+....+a_{2015}\right)=1008\Leftrightarrow a_1+0=1008\) \(\Rightarrow a_1=1008\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có;
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2018}}{a_{2019}}=\frac{a_1+a_2+...+a_{2018}}{a_2+a_3+...+a_{2019}}\)(1)
Ta có:
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2018}}{a_{2019}}\Rightarrow\frac{a_1^{2018}}{a_2^{2018}}=\frac{a_1^{2018}}{a_2^{2018}}=\frac{a_2^{2018}}{a_3^{2018}}=...=\frac{a_{2018}^{2018}}{a_{2019}^{2018}}=\frac{a_1\cdot a_2\cdot...a_{2018}}{a_2\cdot a_3\cdot...\cdot a_{2019}}=\frac{a_1}{a_{2019}}\)(2)
Từ (1) và (2) suy ra
\(\frac{a_1^{2018}}{a_2^{2018}}=\frac{a_2^{2018}}{a_3^{2018}}=...=\frac{a_{2018}^{2018}}{a_{2019}^{2018}}=\left(\frac{a_1+a_2+...+a_{2018}}{a_2+a_3+...+a_{2019}}\right)^{2018}\)(3)
Từ (1), (2), (3) suy ra điều phải chứng minh
\(a_1+a_2+a_3+..+a_{2015}=0\)\(0\)
\(\Rightarrow\left(a_1+a_2\right)+...+\left(a_1+a_{2015}\right)\)\(=\frac{\left(2015-1\right)}{2}+1=1008\)
\(\Rightarrow a_1+\left(a_1+a_2+..+a_{2015}\right)=1008\)
\(\Rightarrow a_1=1008\)
Ta có:
\(a_1+a_2+...+a_{2015}=0\)
\(\Leftrightarrow\left(a_1+a_2\right)+\left(a_3+a_4\right)+...+\left(a_{2013}+a_{2014}\right)+\left(a_{2015}+a_1\right)-a_1=0\)
\(\Leftrightarrow1+1+...+1-a_1=0\)
\(\Leftrightarrow1008-a_1=0\)
\(\Leftrightarrow a_1=1008\)
Ta có:
\(a_2^2=a_1.a_3;a_3^2=a_2.a_4;...;a^2_{2010}=a_{2009}.a_{2011}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3};\frac{a_2}{a_3}=\frac{a_3}{a_4};...;\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=...=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}=\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}\) (1)
Ta lại có:
\(\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_1}{a_2}.\frac{a_1}{a_2}...\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}...\frac{a_{2009}}{a_{2010}}.\frac{a_{2010}}{a_{2011}}=\frac{a_1}{a_{2011}}\) (2)
Từ (1) và (2) ta suy ra
\(\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\)
Ta có :
\(a_2^2=a_1.a_3\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}\)
\(a^2_3=a_2.a_4\Rightarrow\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
\(............\)
\(a^2_{2010}=a_{2009}.a_{2011}\Rightarrow\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=........=\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=.......=\frac{a_{2010}}{a_{2011}}=k\)
\(\Rightarrow a_1=a_2.k\)
\(\Rightarrow a_1=a_3.k^2\)
\(\Rightarrow a_1=a_4.k^3\)
\(...............\)
\(\Rightarrow a_1=a_{2011}.k^{2010}\)
\(\Rightarrow\frac{a_1}{a_{2011}}=k^{2010}\) (1)
Ta có : \(k^{2010}=\left(\frac{a_1}{a_2}\right)^{2010}=\left(\frac{a_2}{a_3}\right)^{2010}=...=\left(\frac{a_{2010}}{a_{2011}}\right)^{2010}=\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=....=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}\)
\(=\frac{a_1^{2010}+a_2^{2010}+a_3^{2010}+....+a^{2010}_{2010}}{a_2^{2010}+a_3^{2010}+a_4^{2010}+....+a_{2011}^{2010}}\) ( theo TC DTSBN ) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a_1^{2010}+a_2^{2010}+....+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+....+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\) (đpcm)
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{100}}{a_1}=\frac{a_1+a_2+...+a_{100}}{a_1+a_2+...+a_{100}}=1\)\(\Rightarrow\)\(a_1=a_2=...=a_{100}\)
\(\Rightarrow\)\(M=\frac{a_1^2+a_2^2+a_3^2+...+a_{100}^2}{\left(a_1+a_2+a_3+...+a_{100}\right)^2}=\frac{100a_1^2}{100^2a_1^2}=\frac{1}{100}\)
Ta có : \(a_1+(a_2+a_3+a_4)+...+(a_{11}+a_{12}+a_{13})+a_{14}+(a_{15}+a_{16}+a_{17})+(a_{18}+a_{19}+a_{20})< 0\)
\(a_1>0;a_2+a_3+a_4>0;....;a_{11}+a_{12}+a_{13}>0;a_{15}+a_{16}+a_{17}>0;a_{18}+a_{19}+a_{20}>0\Rightarrow a_{14}< 0\)
Cũng như vậy : \((a_1+a_2+a_3)+...+(a_{10}+a_{11}+a_{12})+(a_{13}+a_{14})+(a_{15}+a_{16}+a_{17})+(a_{18}+a_{19}+a_{20})< 0\)
\(\Rightarrow a_{13}+a_{14}< 0\)
Mặt khác : \(a_{12}+a_{13}+a_{14}>0\Rightarrow a_{12}>0\)
Từ các điều kiện \(a_1>0;a_{12}>0;a_{14}< 0\Rightarrow a_1\cdot a_{14}+a_{14}\cdot a_{12}< a_1\cdot a_{12}(đpcm)\)
P/S : Hoq chắc :>