已知数列﹛an﹜满足:a1+3a2+……+(2n-1)an=(2n-3)*2^n+1,数列﹛bn﹜的前n项和为Sn=2n

2个回答

  • (2*1-3)*2^1 + 1 = -1 = a(1),

    (2n+2-3)*2^(n+1) - (2n-3)2^n = (2n+2-1)a(n+1) = (2n+1)a(n+1)

    =(2n-1)*2*2^n - (2n-3)2^n

    =(4n-2-2n+3)2^n

    =(2n+1)2^n,

    a(n+1)=2^n,

    a(1)=-1,

    n>=2时,a(n)=2^(n-1).

    b(1)=s(1)=2+1-2=1,

    b(n+1)=s(n+1)-s(n)=2(2n+1)+1=4n+3=4(n+1)-1,

    b(1)=1,

    n>=2时,b(n)=4n-1.

    c(n)=a(n)b(n),

    c(1)=-1,

    n>=2时,c(n)=(4n-1)2^(n-1),

    w(n)=c(1)+c(2)+c(3)+...+c(n)

    =-1 + (4*2-1)2 + (4*3-1)2^2 + ...+ [4(n-1)-1]2^(n-2) + (4n-1)2^(n-1),

    2w(n) = -2 + (4*2-1)2^2 + (4*3-1)2^3 + ...+ [4(n-1)-1]2^(n-1) + (4n-1)2^n,

    w(n) = 2w(n)-w(n) = -1 - (4*2-1)2 - 4*2^2 - ...- 4*2^(n-1) + (4n-1)2^n

    = (4n-1)2^n - 4[1+2+2^2+...+2^(n-1)] + 4(1+2) - 1 - 7*2

    =(4n-1)2^n - 4[2^n - 1]/(2-1) + 12 - 1 - 14

    =(4n-1)2^n - 4[2^n - 1] - 3

    =(4n-5)2^n + 1