本人学渣,

1个回答

  • (1)a(n+1)/an=n/(n+1)

    an/a(n-1)=(n-1)/n

    .

    a2/a1=1/2

    所以a(n+1)/an*an/a(n-1) *.*a2/a1

    =a(n+1)/a1=1/(n+1)

    所以a(n+1)=1/(n+1)

    an=1/n

    (2)sn=n(2n-1)an

    s(n-1)=(n-1)(2n-3)a(n-1)

    两式相减得

    an=sn-s(n-1)=n(2n-1)an -(n-1)(2n-3)a(n-1)

    an*(2n^2-n-1)=a(n-1)*(2n^2-5n+3)

    an/a(n-1)=(2n^2-5n+3)/(2n^2-n-1) =(n-1)(2n-3)/[(2n+1)(n-1)]=(2n-3)/(2n+1)

    a(n-1)/a(n-2)=(2n-5)/(2n-1)

    a(n-2)/a(n-3)=(2n-7)/(2n-3)

    .

    a3/a2=3/7

    a2/a1=1/3

    所以an/a(n-1)*a(n-1)/a(n-2)*a(n-2)/a(n-3)*.*a3/a2*a2/a1

    =an/a1=3/[(2n-1)(2n-3)]

    an=1/[(2n-1)(2n-3)]

    (3)a(n+1)=2/3 an +4

    令a(n+1)-k=2/3 (an-k)

    则k-2/3 k =4

    所以k=12

    a(n+1)-12=2/3 (an-12)

    {a(n+1)-12)}是以公比为2/3 的等比数列

    所以a(n+1)-12=(2/3)^n *(a1-12)

    a(n+1)=12+(2/3)^n *(-11)

    an=12+(2/3)^(n-1)*(-11)