Question

Chứng minh rằng nếu: \(\frac{a_1}{a_2}\)=\(\frac{a_2}{a_3}\)=...=\(\frac{a_n}{a_{n+1}}\)(n \(\varepsilon\)\(ℕ^∗\))

thì:\(\frac{\left(a_1+a_2+...+a_n\right)^n}{\left(a_2+a_3+...+a_{n+1}\right)^n}\)=\(\frac{a^n_1+a^n_2+...+a_n^n}{a^n_2+...+a_{n+1}^n}\)=\(\frac{a_1}{a_{n+1}}\)

 

 

Answer 1

Ta có :

\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)

\(\Rightarrow\)\(\frac{a_1^n}{a_2^n}=\frac{a_2^n}{a_3^n}=...=\frac{a_n^n}{a_{n+1}^n}=\frac{a_1^n+a_2^n+...+a_n^n}{a_2^n+a_3^n+...+a_{n+1}^n}=\frac{\left(a_1+a_2+...+a_n\right)^n}{\left(a_2+a_3+...+a_{n+1}\right)^n}=\frac{a_1.a_2...a_n}{a_2.a_3...a_{n+1}}=\frac{a_1}{a_{n+1}}\)