Tính tổng A =\(\frac{c}{a_1.a_2}+\frac{c}{a_2.a_3}+....+\frac{c}{a_{n-1}.a_n}\)với \(a_2-a_1=a_3-a_2=...=a_n-a_{n-1}=k\)
CMR:
Nếu \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}\)thì\(\left(\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+..+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)
CmR nếu \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}\)
thì\(\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)
Ai giúp tớ đi , nói cách làm thôi cũng được :v
cho \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1};a_1+a_2+..+a_{n-1}+a_n\ne0\)
Tính \(\frac{a^2_2+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)^2}\)
Cho dãy tỉ số bằng nhau \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}\) biết \(a_1+a_2+a_3+...+a_n\ne0;a_1=\sqrt{3}\)
Tính tổng \(a_1+a_2+a_3+...+a_n\)
\(Cho\) \(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=...=\dfrac{a_{n-1}}{a_n}=\dfrac{a_n}{a_1}\). Và \(a_1+a_2+...+a_n\ne0;a_1=-\sqrt{5}\). Tính \(a_2;a_3;...a_n=?\)
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}\)
\(a_1+a_2+...+a_n\ne0;a_1=-\sqrt{5}\)
tính a2;a3;...;an
Cho \(\frac{a_1}{a_2}\)=\(\frac{a_2}{a_3}\)=..........=\(\frac{a_{n-1}}{a_n}\)=\(\frac{a_n}{a_1}\); \(a_1\)+\(a_2\)+.......+\(a_{n-1}\)+\(a_n\)# 0
Tính \(\frac{a_1^2+a_2^2+...+a_n^2}{\left(a_1+a_2+....+a_n\right)^2^{ }}\)
Chứng minh rằng nếu: \(\frac{a_1}{a_2}\)=\(\frac{a_2}{a_3}\)=...=\(\frac{a_n}{a_{n+1}}\)(n \(\varepsilon\)\(ℕ^∗\))
thì:\(\frac{\left(a_1+a_2+...+a_n\right)^n}{\left(a_2+a_3+...+a_{n+1}\right)^n}\)=\(\frac{a^n_1+a^n_2+...+a_n^n}{a^n_2+...+a_{n+1}^n}\)=\(\frac{a_1}{a_{n+1}}\)