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Old 11 ¾ÄÉÀÒ¤Á 2009, 13:12
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\sum_{n = 1}^{\infty}[sin (n\pi - \pi /2) + (-1)¡¡ÓÅѧ n+1 ·Ñé§ËÁ´Êèǹ (-1) ¡¡ÓÅѧ n+2 ¤Ù³ -3]·Ñé§ËÁ´Â¡¡ÓÅѧ n
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11 ¾ÄÉÀÒ¤Á 2009 13:30 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ B º ....
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Old 11 ¾ÄÉÀÒ¤Á 2009, 13:29
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§§¡Ñºâ¨·Âì¤ÃѺẺ¹ÕéäËÁ¤ÃѺ
$$\sum_{n=1}^{\infty } [\dfrac{\sin (n\pi-\frac{\pi}{2})+(-1)^{n+1}}{(-3)(-1)^{n+2}}]^n$$
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Old 11 ¾ÄÉÀÒ¤Á 2009, 13:31
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Old 11 ¾ÄÉÀÒ¤Á 2009, 13:35
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Old 11 ¾ÄÉÀÒ¤Á 2009, 13:38
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Old 11 ¾ÄÉÀÒ¤Á 2009, 13:39
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$$\sum_{n=1}^{\infty } [\dfrac{\sin (n\pi-\frac{\pi}{2})+(-1)^{n+1}}{(-3)(-1)^{n+2}}]^n$$
ÊѧࡵÇèÒ $\sin{\left(n\pi-\frac{\pi}{2}\right)}=(-1)^{n+1}$
$\displaystyle\therefore\sum_{n=1}^{\infty } [\dfrac{\sin (n\pi-\frac{\pi}{2})+(-1)^{n+1}}{(-3)(-1)^{n+2}}]^n=\sum_{n=1}^{\infty } \left(\frac{2}{3}\right)^n=2$
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Old 11 ¾ÄÉÀÒ¤Á 2009, 13:43
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äÁèÂÒ¡ÍÂèÒ§·Õè¤Ô´¤ÃѺ á»Å§ à·ÍÁ·ÕèÁÕ $sine$ ãËéÍÍ¡¡çä´éáÅéÇÅèФÃѺ
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  #8  
Old 11 ¾ÄÉÀÒ¤Á 2009, 13:48
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$$n=1;[\dfrac{\sin (\pi-\frac{\pi}{2})+1}{(-3)(-1)}]^1=\frac{2}{3}$$
$$n=2;[\dfrac{\sin (2\pi-\frac{\pi}{2})-1}{(-3)(1)}]^2=(\frac{2}{3})^2$$
$$n=3;[\dfrac{\sin (3\pi-\frac{\pi}{2})+1}{(-3)(-1)}]^3=(\frac{2}{3})^3$$
$$.$$
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ä´éÇèÒ
$$\sum_{n=1}^{\infty} [\dfrac{\sin (n\pi-\frac{\pi}{2})+(-1)^{n+1}}{(-3)(-1)^{n+2}}]^n=\frac{2}{3}+(\frac{2}{3})^2+(\frac{2}{3})^3+...$$
$$S_{\infty }=\frac{2}{3}+(\frac{2}{3})^2+(\frac{2}{3})^3+...$$
¨Ò¡Êٵà $S_{\infty}=\dfrac{a_1}{1-r}$
ä´éÇèÒ $a_1=\frac{2}{3} $ áÅÐ $r=\frac{2}{3}$
´Ñ§¹Ñé¹
$$S_{\infty }=\frac{2}{3}+(\frac{2}{3})^2+(\frac{2}{3})^3+...=\dfrac{\frac{2}{3}}{1-\frac{2}{3}}=2$$
$$\therefore \sum_{n=1}^{\infty} [\dfrac{\sin (n\pi-\frac{\pi}{2})+(-1)^{n+1}}{(-3)(-1)^{n+2}}]^n=2$$
»Å.äÁè·ÃÒºÇèÒ¶Ù¡äËÁ¹Ð¤ÃѺ
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Old 11 ¾ÄÉÀÒ¤Á 2009, 19:55
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