Question

[Trương Quỳnh Hoa]

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

 Chứng minh rằng ta có đẳng thức: \(\frac{a_1}{a_{2009}}=\left(\frac{a_1+a_2+a_3+...+a_{2008}}{a_2+a_3+a_4+...+a_{2009}}\right)^{2008}\)

Answer 1

[Hoàng Phúc]

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)