数列an的通项公式an=n(cosnπ/3²-sinnπ/3²)其前n项和为Sn,则S30等于多少

数列an的通项公式an=n(cosnπ/3²-sinnπ/3²)其前n项和为Sn,则S30等于多少

（1）∵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(2)∵当n=3k时：S[3k]=n(3n+4)/6=3k(9k+4)/6=k(9k+4)/2∴S[3n]=n(9n+4)/2∵b[n]=S[3n]/(n4^n) ∴b[n]=(9n+4)/(2*4^n)=(9/2)(n/4^n)+2/4^n设R[n]=1/4^1+2/4^2+3/4^3+...+n/4^n则R[n]/4=1/4^2+2/4^3+3/4^4+...+n/4^(n+1)∴3R[n]/4=R[n]-R[n]/4=(1/4^1+1/4^2+1/4^3+...+1/4^n)-n/4^(n+1)=(1/4)(1-1/4^n)/(1-1/4)-n/4^(n+1)=(1/3)(1-1/4^n)-n/4^(n+1)=[4-(3n+4)/4^n]/12∴R[n]=[4-(3n+4)/4^n]/9∴T[n]=[(9/2)(1/4^1)+2/4^1]+[(9/2)(2/4^2)+2/4^2]+...+[(9/2)(n/4^n)+2/4^n]=(9/2)(1/4^1+2/4^2+...+n/4^n)+2(1/4^1+1/4^2+1/4^3+...+1/4^n)=(9/2)R[n]+2(1/4)(1-1/4^n)/(1-1/4)=[4-(3n+4)/4^n]/2+(2/3)(1-1/4^n)=2-(3n/2+2)/4^n+2/3-(2/3)/4^n=[16-(9n+16)/4^n]/6
再问： 这是原式 ...an=n(cosnπ/3²-sinnπ/3²).......你回答的是哪道题啊