1.chung minh rang:
b)$\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+$\frac{100}{3^{100}}$<$\frac{3}{4}$
Mới nghĩ ra. Cách giải của mình chưa thật ngắn gọn.Mong bạn thông cảm
Ta có
$S=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+...+\frac{100}{3^{100}}$
$\Rightarrow \frac{1}{3}S=\frac{1}{3^{2}}+\frac{2}{3^{3}}+\frac{3}{3^{4}}+...+\frac{100}{3^{101}}$
$\Rightarrow \frac{2}{3}S=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}$
$\Rightarrow \frac{2}{9}S=\frac{1}{3^{2}}+\frac{1}{3^{3}}+...+\frac{1}{3^{101}}-\frac{100}{3^{102}}$
$\Rightarrow \frac{4}{9}S=\frac{1}{3}-\frac{1}{3^{101}}+\frac{100}{3^{102}}-\frac{100}{3^{101}}$
$\Rightarrow S=\frac{9}{4}(\frac{1}{3}-\frac{1}{3^{101}}+\frac{100}{3^{102}}-\frac{100}{3^{101}})$
$=\frac{9}{4}.\frac{1}{3}.3(\frac{1}{3}-\frac{1}{3^{101}}+\frac{100}{3^{102}}-\frac{100}{3^{101}})=\frac{3}{4}(1-\frac{1}{3^{100}}+\frac{100}{3^{101}}-\frac{100}{3^{100}})=\frac{3}{4}(1+\frac{100}{3^{101}}-\frac{102}{3^{100}})$
Mà $\frac{100}{3^{101}}< \frac{102}{3^{100}}$
Do vậy $\frac{3}{4}(1+\frac{100}{3^{101}}-\frac{102}{3^{100}})< \frac{3}{4}.1=\frac{3}{4}$
Hay $S< \frac{3}{4}$