在数列{an}中,a1=2,a n+1=4an-3n+1,n∈正整数

1个回答

  • (1)A(n+1)-(n+1)=4An-4n=4(An-n)

    A1-1=1≠0

    ∴{An-n}是等比数列

    (2)An-n=4^(n-1)*(A1-1)=4^(n-1)

    ∴An=n+4^(n-1)

    Sn=A1+A2+A3+……+An

    =1+4^0+2+4^1+3+4^2+……+n+4^(n-1)

    =(1+2+3+……+n)+(1+4+4^2+……+4^(n-1))

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

    =1/3*4^n+1/2*n^2+1/2*n-1/3

    (3)4Sn=1/3*4^(n+1)+2*n^2+2*n-4/3

    S(n+1)=1/3*4^(n+1)+1/2*(n+1)^2+1/2*(n+1)-1/3

    4Sn-S(n+1)=(1/3*4^(n+1)+2*n^2+2*n-4/3)-(1/3*4^(n+1)+1/2*(n+1)^2+1/2*(n+1)-1/3)

    =3/2*n^2+1/2n-2

    =1/2(n-1)(3n+4)

    n≥1

    所以4Sn-S(n+1)=1/2(n-1)(3n+4)≥0

    ∴S(n+1)≤4Sn

    --------------------------------------------------------

    希望可以帮到你!

    如对回答满意,

    --------------------------------------------------------