数列an=n²cosnπ/3求和,谢谢

数列an=n²cosnπ/3求和,谢谢 likang789 1年前他留下的回答 已收到1个回答

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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

1年前他留下的回答

9

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