已知各项都不相等的等差数列{an}的前6项和为60,且a6为a1和a21的等比中项>

1.
设{an}公差为d,各项均不相等,d≠0
S6=6a1+15d=60
a1+(5/2)d=10
a1=10-(5/2)d
a6是a1与a21的等比中项
a6²=a1·a21
(a1+5d)²=a1·(a1+20d)
25d²-10a1d=0
5d(5d-2a1)=0
d≠0,因此2a1=5d a1=(5/2)d
(5/2)d=10-(5/2)d
5d=10
d=2
a1=(5/2)d=5
an=a1+(n-1)d=5+2(n-1)=2n+3
Sn=(a1+an)n/2=(5+2n+3)n/2=n²+4n
数列{an}的通项公式为an=2n+3,前n项和Sn=n²+4n.
2.
b(n+1)-bn=an=2n+3
bn-b(n-1)=2(n-1)+3
b(n-1)-b(n-2)=2(n-2)+3
…………
b2-b1=2+3
累加
bn-b1=2[1+2+...+(n-1)]+3(n-1)=2n(n-1)/2 +3(n-1)=n²+2n-3
bn=b1+n²+2n-3=3+n²+2n-3=n²+2n=n(n+2)
1/bn=1/[n(n+2)]=(1/2)[1/n -1/(n+2)]
Tn=1/b1+1/b2+...+1/bn
=(1/2)[1/1-1/3+1/2-1/4+...+1/n-1/(n+2)]
=(1/2)[(1+1/2+...+1/n)-(1/3+1/4+...+1/(n+2))]
=(1/2)[1+1/2 -1/(n+1)-1/(n+2)]
=3/4 -1/[2(n+1)] -1/[2(n+2)]