Question

Chứng minh: M= \(\dfrac{1}{3}\)+  \(\dfrac{2}{3^2}\)+ \(\dfrac{3}{3^3}\) +  .....+  \(\dfrac{100}{3^{100}}\)  <\(\dfrac{3}{4}\)

Answer 1

Lời giải:

$M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}$

$3M=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}$

$\Rightarrow 2M=3M-M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}$

$2M+\frac{100}{3^{100}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$

$3(2M+\frac{100}{3^{100}})=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}

$\Rightarrow 2(2M+\frac{100}{3^{100}})=3(2M+\frac{100}{3^{100}})-(2M+\frac{100}{3^{100}})=2-\frac{1}{3^{99}}$

$M=\frac{1}{2}-\frac{1}{4.3^{99}}-\frac{50}{3^{100}}<\frac{1}{2}< \frac{3}{4}$ 
Ta có đpcm.