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#1
10 Á¡ÃÒ¤Á 2015, 21:33
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¼ÅºÇ¡Í¹Ñ¹µì
$\frac{A}{(x+c)^2} = \frac{A}{c^2}\sum_{n = 1}^{\infty} {n(\frac{-x}{c})^{n-1}=\frac{A}{c^2}\sum_{n = 0}^{\infty} {(n+1)(\frac{-x}{c})^{n}}}$
â´Â·Õè A áÅÐ c à»ç¹¤èÒ¤§·Õè ¢éͤÇÒÁ¹Õé¶Ù¡äËÁ¤ÃѺ 10 Á¡ÃÒ¤Á 2015 21:45 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ mark123 ^.^ 
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#2
11 Á¡ÃÒ¤Á 2015, 21:07
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$$\frac{A}{c^2}\frac{1}{(1+\frac{x}{c})^2}$$
¾Ô¨ÒÃ³Ò $\frac{1}{1+x}=1-x+x^2-x^3+....$ $\frac{-1}{(1+x)^2}=-1+2x-3x^2+4x^3+....$ $\frac{1}{(1+x)^2}=1-2x+3x^2-4x^3+....$ $$\frac{A}{(x+c)^2}=\frac{A}{c^2}\frac{1}{(1+\frac{x}{c})^2} \\ =\frac{A}{c^2}\bigg(1-2\frac{x}{c}+3\frac{x^2}{c^2}-4\frac{x^3}{c^3}+...\bigg)\\ =\frac{A}{c^2}\sum_{n = 1}^{\infty} n(\frac{-x}{c})^{n-1}$$ ÍÕ¡ÍѹÅͧ¤Ô´´Ù¹Ð¤ÃѺ
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#3
11 Á¡ÃÒ¤Á 2015, 22:52
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ÍéÒ§ÍÔ§:
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