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A.P.O. Akratitos is a
club from , . They currently play in the Alpha Division of the local
championships, which is the fifth level of the
pyramid. They maintain a quite strong academy of youngsters.
Akratitos spent their first years in relative obscurity, until they rose from the lower ranks to the top flight under the managership of , who died on the pitch during a training session in 2002. The stadium was subsequently named after him.
Akratitos spent 3 consecutive seasons (2001-02 until 2003-04) in the
before being relegated in 2004 after losing a relegation match against
which took place in , . The following season Akratitos were re-promoted to Alpha Ethniki.
season was meant to be their last in the Greek top-flight, as they finished last and were relegated again to the , but they decided to withdraw from the professional leagues completely and were automatically demoted to the . During that season they also achieved the lowest attendance ever in the Alpha Ethniki, with only 26 spectators watching their home game against .
The team faced another relegation in 2009, when they finished 12th in Group 7 of the
and got relegated to the local
championships.
In July 2003 Akratitos took part in the . They played against
for the second round, using their youth team, and were eliminated.
Competition
The team is using the
as their home ground. Until the death of Giannis Pathiakakis, it was known as Akratitos Stadium. It has a capacity of 4,944.
Runners-up (2): 2000–01,
(in Greek). erasitexniko-podosfairo.gr 2012.
(in Greek). West Attica Football Clubs Association 2012.
(in Greek). . 9 February 2002.
(in Greek). Insports.gr 2012.
(in Greek). . 30 May 2004.
(in Greek). Patris. 22 July 2006.
Saridakis, Manolis (20 December 2005).
(in Greek). .
(in Greek). Nooz.gr. 20 December 2005.
(in Greek). Dokari.gr. 17 May 2009.
(in Greek). . 15 July 2003.
(in Greek)
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{list wl as x}{/list}已知函数f(x)=x3对应的曲线在点(ak.f(ak))(k∈N*)处的切线与x轴的交点为(ak+1.0).若a1=1.则f(3a1)+f(3a2)+-+f(3a10)1-(23)10= ． 题目和参考答案——精英家教网——
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已知函数f（x）=x3对应的曲线在点（ak，f（ak））（k∈N*）处的切线与x轴的交点为（ak+1，0），若a1=1，则f(3a1)+f(3a2)+…+f(3a10)1-(23)10=．
考点：利用导数研究曲线上某点切线方程
专题：计算题,导数的概念及应用,等差数列与等比数列
分析：求出函数的导数，可得切线的斜率，由点斜式方程可得切线方程，再令y=0，结合等比数列的定义可得，数列{an}是首项a1=1，公比q=23的等比数列，再由等比数列的求和公式计算即可得到所求值．
解：由f'（x）=3x2得曲线的切线的斜率k=3a2k，故切线方程为y-a3k=3a2k(x-ak)，令y=0得ak+1=23ak⇒ak+1ak=23，故数列{an}是首项a1=1，公比q=23的等比数列，又f(3a1)+f(3a2)+…+f(3a10)=a1+a2+…+a10=a1(1-q10)1-q=3(1-q10)，所以f(3a1)+f(3a2)+…+f(3a10)1-(23)10=3．故答案为：3．
点评：本题考查导数的运用：求切线的方程，主要考查导数的几何意义，同时考查等比数列的定义和求和公式，运用点斜式方程求得切线方程是解题的关键．
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