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