Sn为数列{an}的前n项和.a1=1,Sn=nan-2n(n-1)（1）求证数列{an}为等差数列（2）设数列{1\anan1}的前n项和为Tn,求Tn

Question

Sn为数列{an}的前n项和.a1=1,Sn=nan-2n(n-1)
（1）求证数列{an}为等差数列（2）设数列{1\anan1}的前n项和为Tn,求Tn

Answer 1

解
当n>=2时
sn=nan-2n(n-1)=nan-2n²+2n
s(n-1)=(n-1)a(n-1)-2(n-1)(n-2)
=(n-1)a(n-1)-2n²+6n-4
∴
an=nan-(n-1)a(n-1)-4n+4
(n-1)a(n-1)=(n-1)an-4(n-1)
两边除以(n-1)
∴an-a(n-1)=4
又a1=1
∴an是以a1=1为首项,d=4为公差的等差数列
∴an=1+4(n-1)=4n-3
1/ana(n+1)=1/(4n-3)(4n+1)=1/4[1/(4n-3)-1/(4n+1)]
TN=1/4[(1-1/5)+(1/5-1/9)+……+(1/(4n-3)-1/(4n+1))]
=1/4[1-1/(4n+1)]
=1/4[4n/(4n+1)]
=n/(4n+1)