设\(\triangle ABC\)的面积为\(1\).
如图\(1\),分别将\(AC\),\(BC\)边\(2\)等分,\(D_{1}\),\(E_{1}\)是其分点,连接\(AE_{1}\),\(BD_{1}\)交于点\(F_{1}\),得到四边形\(CD_{1}F_{1}E_{1}\),其面积\(S_{1}= \dfrac {1}{3}\).
如图\(2\),分别将\(AC\),\(BC\)边\(3\)等分,\(D_{1}\),\(D_{2}\),\(E_{1}\),\(E_{2}\)是其分点,连接\(AE_{2}\),\(BD_{2}\)交于点\(F_{2}\),得到四边形\(CD_{2}F_{2}E_{2}\),其面积\(S_{2}= \dfrac {1}{6}\);
如图\(3\),分别将\(AC\),\(BC\)边\(4\)等分,\(D_{1}\),\(D_{2}\),\(D_{3}\),\(E_{1}\),\(E_{2}\),\(E_{3}\)是其分点,连接\(AE_{3}\),\(BD_{3}\)交于点\(F_{3}\),得到四边形\(CD_{3}F_{3}E_{3}\),其面积\(S_{3}= \dfrac {1}{10}\);
\(…\)
按照这个规律进行下去,若分别将\(AC\),\(BC\)边\((n+1)\)等分,\(…\),得到四边形\(CD_{n}E_{n}F_{n}\),其面积\(S=\) ______ .