Question

Cho A=\(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)

CMR: A<\(\frac{5}{3}\)

Answer 1

\(A=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\)

\(A=5\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)\)

\(\frac{A}{5}=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)

\(\frac{4A}{5}=1+\frac{1}{4}+...+\frac{1}{4^{98}}\)

\(\frac{4A}{5}-\frac{A}{5}=\left(1+\frac{1}{4}+...+\frac{1}{4^{98}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)\)

\(\frac{3A}{5}=1-\frac{1}{4^{99}}\Rightarrow A=\frac{5}{3}-\frac{5}{3\cdot4^{99}}< \frac{5}{3}\)