Câu hỏi của kiwi nguyễn

Question

CMR:

👁💧👄💧👁 · Accepted Answer

Đặt $A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}$

$\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}$

$3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)$

$2A=1-\frac{1}{3^{99}}$

$\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\left(đpcm\right)$

Câu trả lời:

Đặt $A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}$

$\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}$

$3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)$

$2A=1-\frac{1}{3^{99}}$

$\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\left(đpcm\right)$