Question

Cho S= 1/3+1/3²+1/3³+...+1/3⁹⁹. Hãy so sánh A với 1/2

Chiều nay mình nộp rồi đó là ngắn gọn dễ hiểu thôi nhé!!!

Answer 1

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)

\(3A=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)

\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)

\(2A=1-\frac{1}{3^{99}}\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}=\frac{1}{2}-\frac{\frac{1}{3^{99}}}{2}< \frac{1}{2}\)

Vậy \(A< \frac{1}{2}\)