Question

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

Answer 1

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)

Answer 2

解：
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)