Question

so sanh  

\(\frac{1+5+5^2+......+5^9}{1+5+5^2+......+5^8}\)  va  \(\frac{1+3+3^2+......+3^9}{1+3+3^2+......+3^8}\)

 

Answer 1

Ta đặt  \(A=1+5+5^2+......+5^9\Rightarrow5A=5+5^2+...+5^9+5^{10}\)

\(\Rightarrow4A=5^{10}-1\Rightarrow A=\frac{5^{10}-1}{4}\)

tTương tự \(B=1+5+5^2+......+5^8\Rightarrow B=\frac{5^9-1}{4}\)

\(C=1+3+3^2+......+3^9\Rightarrow C=\frac{3^{10}-1}{3}\)

\(D=1+3+3^2+......+3^8\Rightarrow D=\frac{3^9-1}{3}\)

Vậy \(\frac{A}{B}=\frac{5^{10}-1}{5^9-1}=\frac{5\left(5^9-1\right)+4}{5^9-1}=5+\frac{4}{5^9-1}\)

\(\frac{C}{D}=\frac{3^{10}-1}{3^9-1}=\frac{3\left(3^9-1\right)+3}{3^9-1}=3+\frac{3}{3^9-1}\)

Ta thấy \(\frac{3}{3^9-1}< 1\Rightarrow3+\frac{3}{3^9-1}< 4< 5< 5+\frac{5}{5^9-1}\)

Vậy \(\frac{A}{B}>\frac{C}{D}\)  hay \(\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}>\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)