Question

Cho a_1/a₂=a₂/a₃=a₃/a₄=...=a₂₀₁₆/a₂₀₁₇

Chứng minh rằng a₁/a₂₀₁₇=(a₁+a₂+...+a₂₀₁₆/a₂+a₃+...+a₂₀₁₇)²⁰¹⁶

Answer 1

Ta có :

\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=...=\frac{a2016}{a2017}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)

vì \(\frac{a1}{a2}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\) 

\(\frac{a2}{a3}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)

...

\(\frac{a2016}{a2017}=\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\)
\(\Rightarrow\frac{a1}{a2}.\frac{a2}{a3}.\frac{a3}{a4}...\frac{a2016}{a2017}=\frac{\left(a1+a2+a3+...+a2016\right)^{2016}}{\left(a2+a3+a4+...+a2017\right)^{2016}}\)

\(\Rightarrow\frac{a1}{a2017}=\left(\frac{a1+a2+a3+...+a2016}{a2+a3+a4+...+a2017}\right)^{2016}\)

Answer 2

Ta có a₁/a₂=a₂/a₃=a₃/a₄=...=a₂₀₁₆/a₂₀₁₇

=> a₁/a₂=(a₁+a₂+a₃+...+a₂₀₁₆)

/(a₂+a₃+a₄+...+a₂₀₁₇₎

=> a₁²⁰¹⁶/a₂²⁰¹⁶ =(a₁+a₂+a₃+...+a₂₀₁₆)²⁰¹⁶ /(a₂+a₃+a₄+...+a₂₀₁₇)²⁰¹⁶ (1)

Ta lại có a₁/a₂=a₂/a₃=a₃/a₄=...=a₂₀₁₆/a₂₀₁₇

=> a₁²⁰¹⁶/a₂²⁰¹⁶= a₁/a_2.a₂/a_3.a₃/a_4. ... .a₂₀₁₆/a₂₀₁₇=a₁/a₂₀₁₇ (2)

Từ (1) và (2) => đpcm