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5Â¥: Originally posted by Edstrayer at 2015-11-15 02:03:52
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0404600213

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hylpy: ½ð±Ò+2, Èç¹ûÖØÐ±àһϣ¬¾Í¸üºÃÁË 2015-11-15 19:15:57
Edstrayer: ½ð±Ò+5, LatexÔ´Îļþ±àÒëͨ¹ý£¬µ«²»·ûºÏСľ³æÂÛ̳µÄ¸ñʽ£¬¹ÄÀøÒ»Ï 2015-11-16 17:25:39
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GOOD NIGHT


\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\par{$$\because 0<a_{1}<7,0<a_{2}<7$$}
\par{$$\because a_{n+2}=\sqrt{7-\sqrt{7+a_{n}}}$$}
\par{$$\therefore 0<a_{2n+1}<7,0<a_{2n+1}<7$$}
\par{$$\therefore 0<a_{n}<7$$}
\par{$$\therefore \{a_{n}\} is\quad bounded.$$}
\par{quad}
\par{$$\because a_{n+2}=\sqrt{7-\sqrt{7+a_{n}}}$$}
\par{$$\therefore a_{4n+i}-a_{4(n-1)+i}=\sqrt{7-\sqrt{7+a_{4n+i-2}}}-\sqrt{7-\sqrt{7+a_{4(n-1)+i-2}}}=\frac{\sqrt{7+a_{4(n-1)+i-2}}-\sqrt{7+a_{4n+i-2}}}{a_{4n+i}+a_{4(n-1)+i}},i=0,1,2,3$$}
\par{$$\therefore a_{4n+i}-a_{4(n-1)+i}=k(a_{4(n-1)+i}-a_{4(n-2)+i}),k>0;i=0,1,2,3$$}
\par{$$\therefore \{a_{4n+i}\} is\quad monotone\quad decreasing.(i=0,1,2,3)$$}
\par{$$\therefore \{a_{4n+i}\} is\quad Convergent\quad sequence.(i=0,1,2,3)$$}
\par{\quad}
\par{$$Suppose\quad \lim_{n\rightarrow\infty} a_{4n+i}=A$$}
\par{$$\because a_{n+2}=\sqrt{7-\sqrt{7+a_{n}}}$$}
\par{$$\therefore A=\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+A}}}}$$}
\par{$$\therefore A=-3,2,\frac{1\pm\sqrt{29}}{2} ......$$}
\par{$$\because 0\leq a_{n}\leq\sqrt{7}$$}
\par{$$\therefore 0\leq A\leq\sqrt{7}$$}
\par{$$\therefore A=2$$}
\par{.....}
\end{document}
2Â¥2015-11-14 23:42:01
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

0404600213

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¡ï
Сľ³æ: ½ð±Ò+0.5, ¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
ÒýÓûØÌû:
2Â¥: Originally posted by 0404600213 at 2015-11-14 23:42:01
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GOOD NIGHT


\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
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3Â¥2015-11-15 00:41:42
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