已知数列{an}的前n项和为Sn,且Sn=(n^2+n-6)/2(n属于N*) (1)求an,(2)设bn=1/(an·

1个回答

  • (1)

    Sn=(n^2+n-6)/2(n属于N*)

    n=1时,a1=-2

    n≥2时,an=Sn-S(n-1)

    =(n²+n-6)/2-[(n-1)²+(n-1)-6]/2

    =1/2*[2n-1+1]=n

    ∴an分分段公式

    an={ -2 ,(n=1)

    {n , (n≥2)

    (2)

    b1=1/(a1a2+1)=-1/3

    n≥2时,bn=1/[n(n+1)+n]

    =1/[n(n+2)]

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

    n=1时,T1=b1=-1/3

    n≥2时,Tn=-1/3+1/2[(1/2-1/4)+(1/3-1/5)+.+1/(n-1)-1/(n+1)+(1/n-1/(n+2))]

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

    =-1/3+1/2[5/6-(2n+3)/(n²+3n+2)]

    =1/12-(2n+3)/(2n²+6n+4)

    n=1时,上式也成立

    ∴Tn= 1/12-(2n+3)/(2n²+6n+4) (n∈N*)