Question

Cho A = 1 + 4 + \(4^2+...+4^{99}\) và B = \(4^{100}\)

Chứng minh rằng \(\frac{A}{B}\) < \(\frac{1}{3}\)

Answer 1

\(4A=4+4^2+...+4^{100}\)

\(A=1+4+4^2+..+4^{99}\)

\(\Rightarrow3A=4A-A=4^{100}-1\)

\(\Rightarrow3A< 4^{100}\)

\(\Rightarrow\frac{3A}{B}< 1\Rightarrow\frac{A}{B}< \frac{1}{3}\)