Question

sos mn

```
b) Cho \( A = \frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + \frac{4}{4^4} + \cdots + \frac{2014}{4^{2014}} \)

Chứng minh \( A < \frac{1}{2} \)
```

Answer 1

a) \(\overline{17xy}⋮42=2.3.7\)

\(\Rightarrow\overline{17xy}⋮\left\{2;3;7\right\}\)

\(\overline{17xy}⋮2\Rightarrow y\in\left\{0;2;4;6;8\right\}\)

\(\overline{17xy}⋮3\Rightarrow1+7+x+y=8+x+y⋮3\)

Thử các giá trị \(\left(x;y\right)\) ta thấy \(\left(x;y\right)=\left(6;4\right)\Rightarrow\overline{17xy}⋮\left\{2;3;7\right\}=1764\)

b) \(...\Rightarrow\dfrac{1}{4}A=\dfrac{1}{4^2}+\dfrac{2}{4^3}+\dfrac{3}{4^4}+...+\dfrac{2014}{4^{2015}}\)

\(\Rightarrow A-\dfrac{1}{4}A=\dfrac{3}{4}A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2014}}-\dfrac{2014}{4^{2015}}\left(1\right)\)

Làm tương tự ta được :

\(\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2014}}=\dfrac{1-\left(\dfrac{1}{4}\right)^{2024}}{3}\)

\(\left(1\right)\Rightarrow\dfrac{3}{4}A=\dfrac{1-\left(\dfrac{1}{4}\right)^{2024}}{3}-\dfrac{2014}{4^{2015}}< \dfrac{1-\left(\dfrac{1}{4}\right)^{2024}}{3}< \dfrac{1}{3}\)

\(\Rightarrow A< \dfrac{1}{3}.\dfrac{4}{3}=\dfrac{4}{9}< \dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{2}\left(đpcm\right)\)

Answer 2

giúp mình câu 11 a và b luôn ạ ! Mình xin cảm ơn trước