一道数列题:在数列{an}中 ,a1=1,2A(n+1)=(1+1/n)2an.求通项公式.

3个回答

  • (1)2a(n+1)=(1+1/n)² an

    a(n+1)/an=1/2 *(n+1)²/n²

    n≥2

    a2/a1=4/1 *1/2

    a3/a2=9/4*1/2

    a4/a3=16/9*1/2

    .

    an/a(n-1)=n^2/(n-1)^2 *1/2

    an/a1=n^2*(1/2)^(n-1)

    an=n^2*(1/2)^(n-1)

    n=1时,上式成立

    所以,an=n^2*(1/2)^(n-1) (n∈N*)

    (2)bn=(n+1)^2/2^n -1/2 *n^2*(1/2)^(n-1)=(2n+1)/2^n

    Sn=3/2+5/4+7/8+.+(2n+1)/2^n 1)

    1/2Sn=3/4+5/8+.+(2n-1)/2^n+(2n+1)/2^(n+1) 2)

    1)-2):

    1/2Sn=3/2+2/4+2/8+.+2/2^n-(2n+1)/2^(n+1)

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

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

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

    Sn=5-(2n+5)/2^n

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

    ∵bn=a(n+1)-1/2an

    两边求和左边为Sn,右边为a2+a3+...+a(n+1)-1/2Tn

    Sn=T(n+1)-1/2 Tn==>Sn=Tn-a1+a(n+1)-1/2Tn

    ==>Tn=2Sn+2-2a(n+1)=12-(2n+5)/2^(n-1)-2(n+1)^2*(1/2)^n=12-(n^2+4n+6)/2^(n-1)