Question

giải hộ với mn :

 

Answer 1

Lời giải:

\(A=\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{99}}\)

$4A=1+\frac{1}{4}+....+\frac{1}{4^{98}}$

$\Rightarrow 4A-A=(1+\frac{1}{4}+....+\frac{1}{4^{98}})-(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}})$

$\Rightarrow 3A=1-\frac{1}{4^{99}}< 1$

$\Rightarrow A< \frac{1}{3}$