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*)
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