Question

[Why Not Me]

Cho biểu thức D =\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\) chứng minh rằng D < \(\frac{3}{4}\)

Answer 1

[Song Hye Hyo Song Joong Ki]

D=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)

D=\(\frac{1}{3}+\frac{101}{3^{101}}\)

D=\(\frac{1}{3}\)

\(\frac{1}{3}và\frac{3}{4}\)

\(\frac{1}{3}=\frac{4}{12}\)

\(\frac{3}{4}=\frac{9}{12}\)

Vì\(\frac{4}{12}< \frac{9}{12}Vậy\frac{1}{3}< \frac{3}{4}\)