![]() |
|
ÊÁѤÃÊÁÒªÔ¡ | ¤ÙèÁ×Í¡ÒÃãªé | ÃÒª×èÍÊÁÒªÔ¡ | »¯Ô·Ô¹ | ¢éͤÇÒÁÇѹ¹Õé | ¤é¹ËÒ |
![]() ![]() |
|
à¤Ã×èͧÁ×ͧ͢ËÑÇ¢éÍ | ¤é¹ËÒã¹ËÑÇ¢é͹Õé |
#1
|
||||
|
||||
![]() ¼Á¤Ô´â¨·ÂìµÍ¹àÃÕ¹ÀÒÉÒä·ÂÅͧ¨ÔéÁã¹ wolfram ¤ÓµÍºÁѹäÁè¹èÒà¡ÅÕ´ÁÒ¡ àŨÐÁÒÅͧ¶ÒÁÇÔ¸Õ¤Ô´¢Í§¤¹ã¹¹Õé´Ù¤ÃѺ
$$\sum_{n = 1}^{\infty} \frac{1}{2^n(n+1)(n+2)} $$ $$\sum_{n = 1}^{\infty}\frac{1}{2^n}\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)$$ $$\sum_{n = 1}^{\infty}\bigg(\frac{1}{2^n(n+1)}-\frac{1}{2^n(n+2)}\bigg)$$ $$\sum_{n = 1}^{\infty}\bigg(\frac{1}{2^n(n+1)}-\frac{1}{2^{n+1}(n+2)}\bigg)-\frac{1}{2^{n+1}(n+2)}$$ $$\frac{1}{4}-\sum_{n = 1}^{\infty}\frac{1}{2^{n+1}(n+2)}$$ µÑÇ¢éÒ§ËÅѧ¹ÕèËÒÂѧä§ËÃͤÃѺ
__________________
Hope is what makes us strong. It's why we are here. It is what we fight with when all else is lost. 25 Á¡ÃÒ¤Á 2015 09:35 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ FranceZii Siriseth |
#2
|
|||
|
|||
![]() ÍéÒ§ÍÔ§:
$-\ln(1-x) = x + \dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots$ àÁ×èÍ $-1< x < 1$ Å᷹ͧ $x=\dfrac{1}{2}$ áÅéǨѴ¾¨¹ìãËéµÃ§¡Ñ¹¡ç¨Ðä´é¤ÓµÍº¤ÃѺ
__________________
site:mathcenter.net ¤Ó¤é¹ |
#3
|
||||
|
||||
![]() ¢Íº¤Ø³ÁÒ¡¤ÃѺ ÍÒ¨ÒÃÂì nooonuii
__________________
Hope is what makes us strong. It's why we are here. It is what we fight with when all else is lost. |
![]() ![]() |
|
|