Question

C/m rằng:

\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)

Answer 1

Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

=> \(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

=> \(3A-A=2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

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

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

=> \(2B=3-\frac{1}{3^{99}}\)

Ta có \(4A=2B-\frac{200}{3^{100}}\)

                = \(3-\frac{1}{3^{99}}-\frac{200}{3^{100}}< 3\)

=> A \(< \frac{3}{4}\)