Question

设数列{an}的前n项和为Sn,已知A1=1,sn=na1-n(n-1),求证数列an为等差数列

Answer 1

(1)对于n>=2,an=sn-s(n-1) = na1-n(n-1)-[(n-1)a1-(n-1)(n-2)]=a1-2(n-1)

a2= a1-2=-1,则a2-a1=-2

an-a(n-1)=a1-2(n-1)-a1+2(n-2)=-2,当n>=3

故an为等差数列.

(2)

由通项公式得an=3-2n

Tn = -1 +1/(1*3)+...+1/(2n-3)(2n-1) = -1 +1/2[1-1/3] +1/2 [1/(2n-3)-1/(2n-1)]=-n/(2n-1)

(不知题中“求满足100/209的最小正整数n”什么意思?)