数列an的通项an=n²（cos²nπ／3－sin²nπ／3）,其前n项的和为sn,则Sn

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数列an的通项an=n²（cos²nπ／3－sin²nπ／3）,其前n项的和为sn,则Sn=?

∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为S[n]
∴a[n]=n^2Cos(2nπ/3)
∴S[n]=1^2(-1/2)+2^2(-1/2)+3^2+4^2(-1/2)+5^2(-1/2)+6^2+...+n^2Cos(2nπ/3)
当n=3k-2,即：k=(n+2)/3时：
S[3k-2]
=(-1/2)[1^2+2^2+...+(3k-2)^2]+3^3/2[1^2+2^2+...+(k-1)^2]
=-(3k-2)(3k-1)(6k-3)/12+27(k-1)k(2k-1)/12
=[-n(n+1)(2n+1)+(n-1)(n+2)(2n+1)]/12
=(2n+1)(-n^2-n+n^2+n-2)/12
=-(2n+1)/6
当n=3k-1,即：k=(n+1)/3时：
S[3k-1]
=(-1/2)[1^2+2^2+...+(3k-1)^2]+3^3/2[1^2+2^2+...+(k-1)^2]
=-(3k-1)3k(6k-1)/12+27(k-1)k(2k-1)/12
=[-n(n+1)(2n+1)+(n-2)(n+1)(2n-1)]/12
=(n+1)(-2n^2-n+2n^2-5n+2)/12
=(n+1)(-6n+2)/12
=-(n+1)(3n-1)/6
当n=3k,即：k=n/3时：
S[3k]
=(-1/2)[1^2+2^2+...+(3k)^2]+3^3/2(1^2+2^2+...+k^2)
=-3k(3k+1)(6k+1)/12+27k(k+1)(2k+1)/12
=[-n(n+1)(2n+1)+n(n+3)(2n+3)]/12
=n(-2n^2-3n-1+2n^2+9n+9)/12
=n(6n+8)/12
=n(3n+4)/6

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