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

问题描述:

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

解当n>=2时sn=nan-2n(n-1)=nan-2n²+2ns(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=...