数列an=n²cosnπ/3求和,谢谢

1个回答

  • an= n^2.cos(nπ/3)

    bn = cos(nπ/3)

    b1 = 1/2

    b2 = -1/2

    b3 = -1

    b4 = -1/2

    b5 = 1/2

    b6 = 1

    .

    .

    an =(1/2)n^2 ; n=1,7,13,...

    =-(1/2)n^2 ; n=2,8,14,...

    =-n^2 ; n=3,9,15,...

    =-(1/2)n^2 ; n=4,10,16,...

    =(1/2)n^2 ; n=5,11,17,...

    =n^2 ; n=6,12,18,.

    bk=a(6k-5)+a(6k-4)+a(6k-3)+a(6k-2)+a(6k-1)+a(6k)

    =(1/2)(6k-5)^2 -(1/2)(6k-4)^2 -(6k-3)^2-(1/2)(6k-2)^2+(1/2)(6k-1)^2 + (6k)^2

    =(1/2)[(6k-5)^2 -(6k-4)^2] -(1/2)[(6k-2)^2-(6k-1)^2] + [(6k)^2-(6k-3)^2]

    =-(1/2)(12k-9) +(1/2)(12k-3)+ 3(2k-3)

    =6k-6

    if n= 6,12,18,...

    a1+a2+...+an

    = b1+b2+...+b(n/6)

    = (n-6)n/12

    if n=5,11,17,...

    a1+a2+...+an

    =[a1+a2+...+an+a(n+1)] - a(n+1)

    =b1+b2+...+b((n+1)/6) - a(n+1)

    =(n-5)(n+1)/6 - (n+1)^2

    if n=4,10,16,...

    a1+a2+...+an

    =[a1+a2+...+a(n+2)] - a(n+1)-a(n+2)

    =b1+b2+...+b((n+2)/6) -a(n+1)-a(n+2)

    =(n-4)(n+2)/12 - (1/2)(n+1)^2-(n+2)^2

    if n=3,9,15,...

    a1+a2+...+an

    =[a1+a2+...+a(n+3)] - a(n+1)-a(n+2)-a(n+3)

    =b1+b2+...+b((n+3)/6) -a(n+1)-a(n+2)-a(n+3)

    =(n-3)(n+3)/12 + (1/2)(n+1)^2-(1/2)(n+2)^2-(n+3)^2

    if n=2,8,14,...

    a1+a2+...+an

    =[a1+a2+...+a(n+4)] - a(n+1)-a(n+2)-a(n+3)-a(n+4)

    =b1+b2+...+b((n+4)/6) -a(n+1)-a(n+2)-a(n+3)-a(n+4)

    =(n-2)(n+4)/12 +(n+1)^2 +(1/2)(n+2)^2-(1/2)(n+3)^2-(n+4)^2

    if n=1,7,13,...

    a1+a2+...+an

    =[a1+a2+...+a(n+5)] - a(n+1)-a(n+2)-a(n+3)-a(n+4)-a(n+5)

    =b1+b2+...+b((n+5)/6) -a(n+1)-a(n+2)-a(n+3)-a(n+4)-a(n+5)

    =(n-1)(n+5)/12 +(1/2)(n+1)^2+(n+2)^2+(1/2)(n+3)^2-(1/2)(n+4)^2-(n+5)^2