Question

Chứng minh rằng 1/3 + 2/3^2 + 3/3^3 + .....+ 100/3^100 < 3/4?

Answer 1

Đặt A = \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^4}...+\dfrac{100}{3^{100}}\)

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

\(\rightarrow2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

6A = \(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\rightarrow4A=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)

4A = \(3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+ \dfrac{100}{3^{100}}\)

4A = 3 - \(\dfrac{203}{3^{100}}\) < 3

\(\Rightarrow\) A < \(\dfrac{3}{4}\) ( đpcm )