Question

[Nguyễn Tiến Đạt]

Cho 2017 số khác 0:a₁,a₂,a₃,....,a₂₀₁₇ thỏa mãn: a₂²=a₁.a_3; a₃²=a₂.a_4;......;a₂₀₁₆²=a₂₀₁₅.a₂₀₁₇

CM/R : \(\frac{a^{2016}_1+a_2^{2016}+a_3^{2016}+....+a_{2016}^{2016}}{a_2^{2016}+a_3^{2016}+a_4^{2016}+....+a_{2017}^{2016}}=\frac{a_1}{a_{2017}}\)

Answer 1

[alibaba nguyễn]

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) \(\frac{a_1^{2016}+a_2^{2016}+...+a_{2016}^{2016}}{a_2^{2016}+a_3^{2016}+...+a_{2017}^{2016}}=\frac{a_1}{a_{2017}}\)