Ta có \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2020}}{a_{2021}}=\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\)(dãy tỉ só bằng nhau)

=> \(\frac{a_1}{a_2}=\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\)

<=>  \(\left(\frac{a_1}{a_2}\right)^{2020}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)

<=> \(\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}...\frac{a_1}{a_2}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)

<=> \(\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}...\frac{a_{2020}}{a_{2021}}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\) 

<=> \(\frac{a_1}{a_{2021}}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)  

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

ta có:

\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{2008}}{a_{2009}}=\frac{a_1+a_2+...+a_{2008}}{a_2+a_3+...+a_{2009}}\)

=>\(\left(\frac{a_1}{a_2}\right)^{2008}=\left(\frac{a_2}{a_3}\right)^{2008}=....=\left(\frac{a_{2008}}{a_{2009}}\right)=\left(\frac{a_1+a_2+..+a_{2008}}{a_2+a_3+..+a_{2009}}\right)^{2008}\)

\(=\frac{a_1a_2}{a_2a_3}=...=\frac{a_{2009}}{a_{2009}}=\frac{a_1}{a_{2009}}\)

=>\(\frac{a_1}{a_{2009}}=\left(....\right)\) (đpcm)

lưu linh ly ơi kết bn với mk nhé

Có: \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=.....=\frac{a_{2008}}{a_{2009}}=\frac{a_1+a_2+a_3+...+a_{2008}}{a_2+a_3+a_4+....+a_{2009}}\)(tính chất dãy tỉ số bằng nhau)

=> \(\left(\frac{a_1}{a_2}\right)^{2008}=\left(\frac{a_2}{a_3}\right)^{2008}=...=\left(\frac{a_{2008}}{a_{2009}}\right)^{2008}=\left(\frac{a_1+a_2+...+a_{2008}}{a_2+a_3+...+a_{2009}}\right)^{2008}\)

\(=\frac{a_1.a_2.....a_{2008}}{a_2.a_3.....a_{2009}}=\frac{a_1}{a_{2009}}\)

=> \(\frac{a_1}{a_{2009}}=\left(\frac{a_1+a_2+...+a_{2008}}{a_2+a_3+....+a_{2009}}\right)^{2008}\)

=> Đpcm

Ta có:

\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=...=\frac{a2008}{a2009}=\frac{\left(a1+a2+...+a2008\right)}{\left(a2+a3+...+a2009\right)}\)

\(\Rightarrow\left(\frac{a1}{a2}\right)^{2008}=\left(\frac{a2}{a3}\right)^{2008}=..=\left(\frac{a2008}{a2009}\right)^{2008}=\left(\frac{a1+a2+..+a2008}{a2+a3+..+a2009}\right)^{2008}\)

\(\Rightarrow\frac{a1.a2....a2008}{a2.a3...a2009}=\left(\frac{a1+a2+..+a2008}{a2+a3+..+a2009}\right)^{2008}\)

\(\Rightarrow\frac{a1}{a2009}=\left(\frac{a1+a2+..+a2008}{a2+a3+..+a2009}\right)^{2008}\)