Question

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.

Answer 1

【1】解：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。

Answer 2

(1)an=n^2cos2πn/3
cos2πn/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n=3k时，Sn=1/2n^2+2/3n
n=3k+1时，Sn=S(n-1)+an=1/2(n-1)^2+2/3(n-1)-1/2n^2=-1/3n-7/6
n=3k+2时，Sn=S(n+1)-a(n+1)=1/2(n+1)^2+2/3(n+1)-(n+1)^2=-1/2n^2-1/3n+1/6
(2)S(3n)=9/2n^2+2n
题目不清，请追问。