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·
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(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*)

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