作业帮已知数列{an}的通项公式an=ncos(nπ/3),其前n项和为Sn,则S2014等于
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已知数列{an}的通项公式an=ncos(nπ/3),其前n项和为Sn,则S2014等于
已知数列{an}的通项公式an=ncos(nπ/3),其前n项和为Sn,则S2014等于
已知数列{an}的通项公式an=ncos(nπ/3),其前n项和为Sn,则S2014等于
函数f(n) = cos(nπ/3)的最小正周期是2π/(π/3) = 6 ； 由已知a n = ncos(nπ/3) ,计算可得：a 1 = cos(π/3) =1*(1/2) = 1/2 ； a 2 = 2cos(2π/3) = 2*(-1/2) = -1 ； a 3 = 3cos(3π/3) =3*(-1) = -3 ； a 4 = 4cos(4π/3) = 4*(-1/2) = -2 ； a 5 = 5cos(5π/3) = 5*(1/2) = 5/2 ； a 6 = 6cos(6π/3) = 6*1 = 6 ； a 7 = 7cos(7π/3) = 7*(1/2) = 7/2 ； …… 所以每隔6项求和有一定规律,考虑到2014/6 = 335……4 ,计算S 2010 = a 1 + a 2 + a 3 + …… + a 2010 = (a 1 + a 7 + …… + a 2005 ) + (a 2 + a 8 + …… + a 2006 ) + (a 3 + a 9 + …… + a 2007 )+ (a 4 + a 10 + …… + a 2008 ) + (a 5 + a 11 + …… + a 2009 ) + (a 6 + a 12 +…… + a 2010 )= (1/2)(1 + 7 + …… + 2005) + (-1/2)(2 + 8 + …… + 2006) + (-1)(3+ 9 + …… + 2007) + (-1/2)(4+ 10 + …… + 2008) + (1/2)(5+ 11 + …… + 2009) + 1(6+ 12 + …… + 2010) = (1/2)[(1+ 7 + …… + 2005) + (5+ 11 + …… + 2009) – (2+ 8 + …… + 2006) – (4+ 10 + …… + 2008)] +[(6 + 12 + …… + 2010) – (3+ 9 + …… + 2007)] =(1/2)[1*335 + (-1)*335] + 3*335 = 1005 ； 而且a 2011 = 2011cos(2011π/3) = 2011/2 ； a 2012 = 2012cos(2012π/3) = -1006 ； a 2013 = 2013cos(2013π/3) = -2013 ； a 2014 = 2014cos(2014π/3) = -1007 ； 所以,S 2014 = S 2010 + a 2011 + a 2012 + a 2013 + a 2014 = 1005 + 2011/2 – 1006 – 2013 – 1007 = - 4031/2 .综上所述,S 2014 = -4031/2 .