数列{an}的通项an=n²(cos²nπ/3-sin²nπ/3),其前n项和为Sn.(1)求Sn;_数学解答

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.

人气：407 ℃　时间：2020-03-29 14:46:41

解答

(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...