Question

cho bieu thuc:A=1/3^2+1/3^3+1/3^4+...+1/3^98+1/3^99

chung to A<1/2

 

Answer 1

ta tính được A=1/6 => 1/6<1/2

Answer 2

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

\(3A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

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

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

\(A=\frac{1}{6}-\frac{1}{2\times3^{99}}\)

Vì \(\frac{1}{2\times3^{99}}>0\) nên \(\frac{1}{6}-\frac{1}{2\times3^{99}}

Answer 3

Đỗ Văn Hoài Tuân trả lời đúng rùi.