数列{an}的n项和为Sn的前n项和Sn=3n^2-2n,bn=3/an*an+1求{bn}的前n项

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数列{an}的n项和为Sn的前n项和Sn=3n^2-2n,bn=3/an*an+1求{bn}的前n项

数列{an}的n项和为Sn的前n项和Sn=3n^2-2n,bn=3/an*an+1求{bn}的前n项
数列{an}的n项和为Sn的前n项和Sn=3n^2-2n,bn=3/an*an+1求{bn}的前n项

数列{an}的n项和为Sn的前n项和Sn=3n^2-2n,bn=3/an*an+1求{bn}的前n项
Sn=3n²-2n,则:
当n=1时,a2=S1=1;
当n≥2时,an=Sn-S(n-1)=6n-5
因n=1也满足,则:an=6n-5
bn=3/[an(an+1)]=3/[(6n-5)(6n+1)]=(1/2)[1/(6n-5)-1/(6n+1)]
则数列{bn}的前n项和Tn=(1/2)[1/1-1/(6n+1)]=(3n)/(6n+1)


n=1
a1=S1=3-2=1
n≥2
an=Sn-S(n-1)
=[3n²-2n]-[3(n-1)²-2(n-1)]
=6n-5
n=1也满足
所以 an=6n-5
bn=3/[(6n-5)(6n+1)]=(1/2)[1/(6n-5)-1/(6n+1)]
Sn=(1/2)[1-1/7+1/7-1/13+.......+1/(6n-5)-1/(6n+1)]
=(1/2)[1-1/(6n+1)]
=3n/(6n+1)