各项均为正数的数列{an},其前n项的和为Sn,

3个回答

  • Sn=(√S(n-1)+√a1)^2

    =S(n-1)+2√a1S(n-1)+a1^2

    Sn-S(n-1)=2√a1S(n-1)+a1^2=an

    2√a1S(n-1)=an-a1^2

    4a1S(n-1)=an^2-2a1an+a1^4

    4a1S(n-2)=a(n-1)^2-2a1a(n-1)+a1^4

    4a1[S(n-1)-S(n-2)]=an^2-a(n-1)^2-2a1[an-a(n-1)]

    4a1a(n-1)=an^2-a(n-1)^2-2a1an+2a1a(n-1)

    an^2-a(n-1)^2-2a1an-2a1a(n-1)=0

    [an+a(n+1)][an-a(n-1)]-2a1[an+a(n+1)]=0

    [an+a(n+1)][an-a(n-1)-2a1]=0

    ∵各项均为正数

    ∴an-a(n-1)-2a1=0

    an-a(n-1)=2a1

    a(n-1)-a(n-2)=2a1

    .

    a2-a1=2a1

    an-a1=2(a1+a1+.a1)

    =2(n-1)a1

    =2a1n-2a1

    an=2a1n-a1

    =a1(2n-1)

    bn=a(n+1)/an+an/a(n+1)

    =a1(2(n+1)-1)/a1(2n-1) +a1(2n-1)/a1(2(n+1)-1)

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

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

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

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

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

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

    Tn=b1+b2+.bn

    =2[1+1-1/3+1+1/3-1/5+.+1+1/(2n-1)-1/(2n+1)]

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

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

    =2[n+2n/(2n+1)]

    =(4n^2+6n)/(2n+1)