数列{an}的通项an=n²(cos²nπ/3-sin²nπ/3),其前n项和为Sn.(1)求Sn;(2)令bn=S(3n) / (n·4n),求数列{bn}的前n项和Tn.
问题描述：
数列{an}的通项an=n²(cos²nπ/3-sin²nπ/3),其前n项和为Sn.
(1)求Sn;
(2)令bn=S(3n) / (n·4n),求数列{bn}的前n项和Tn.
答
【1】an=n²（cos²nπ／3－sin²nπ／3）=n^2*cos(2nπ/3)(二倍角公式)
cos(2π/3)=-1/2
cos(4π/3)=-1/2
cos(6π/3)=1
所以a(3k-2)+a(3k-1)+a(3k)
=(3k-2)^2*(-1/2)+(3k-1)^2*(-1/2)+(3k)^2*1
=9k-5/2
所以S30=a1+a2+...+a30
=(a1+a2+a3)+(a4+a5+a6)+...+(a28+a29+a30)
=(9*1-5/2)+(9*2-5/2)+...+(9*10-5/2)
=9*(1+2+...+10)-10*5/2
=9*10*11/2-25
=470
(2) S(3n)=9/2n^2+2n
题目不清，请追问。
答
(1)an=n^2cos2πn/3cos2πn/3取到的值为-1/2,-1/2,1,-1/2,-1/2,1,.对于n=3k(k∈N*),a(3k-2)+a(3k-1)+a(3k)=-1/2(3k-2)^2-1/2(3k-1)^2+(3k)^2=9k-5/2所以S(3k)为{9k-5/2}的前k项和S(3k)=9k(k+1)/2-5/2k=9/2k^2+2k即当...