回復 9# g112 的帖子
填充3.
如下圖，\(O(0,0)\)，\(A_1(8,0)\)，\(\overline{A_1A_2}\)與\(x\)軸正向夾\(45^{\circ}\)角，又\(\overline{A_1A_2}// \overline{A_3A_4}// \overline{A_5A_6}// \ldots\)，且\(\overline{OA_1}// \overline{A_2A_3}// \overline{A_4A_5}// \ldots\)，已知\(\overline{A_1A_2}=8\)，\(\overline{A_1A_2}=2\overline{A_2A_3}\)，\(\overline{A_2A_3}=2\overline{A_3A_4}\)，\(\ldots\)，\(\overline{A_kA_{k+1}}=2\overline{A_{k+1}A_{k+2}}\)，\(k \in N\)。若點\(A_n\)的坐標為\((x_n,y_n)\)，則\(\displaystyle \lim_{n \to \infty}(x_n+y_n)=\) 。
[解]
\(\matrix{\displaystyle y_n=&4\sqrt{2}&-&\sqrt{2}&+&\frac{\sqrt{2}}{4}&-&\frac{\sqrt{2}}{16}&+\ldots \cr
&A_2&&A_4&&A_6&&A_8&}\)
\(\displaystyle \lim_{n \to \infty}y_n=\frac{4\sqrt{2}}{1-(-\frac{1}{4})}=\frac{16\sqrt{2}}{5}\)
\(\matrix{\displaystyle x_n&8&+&4\sqrt{2}&-&4&-&\sqrt{2}&+&1&+&\frac{\sqrt{2}}{4}&-&\frac{1}{4}&+\ldots \cr
&A_1&&A_2&&A_3&&A_4&&A_5&&A_6&&A_7&}\)
\( x_{n} \)拆成兩個無窮等比級數和。
\(\displaystyle x_n=8+\left(4\sqrt{2}-\sqrt{2}+\frac{\sqrt{2}}{4}-\ldots \right)+\left(-4+1-\frac{1}{4}+\ldots \right)\)
\( \displaystyle \lim_{n \to \infty}x_n=8+\frac{4\sqrt{2}}{1-(-\frac{1}{4})}+\frac{-4}{1-(-\frac{1}{4})}=8+\frac{16\sqrt{2}}{5}-\frac{16}{5}=\frac{24+16\sqrt{2}}{5} \)
\(\displaystyle \lim_{n \to \infty}(x_n+y_n)=\frac{24+16\sqrt{2}}{5}+\frac{16\sqrt{2}}{5}=\frac{24+32\sqrt{2}}{5}\)
