已知数列的前n项和为sn,且sn=2n^2-3n=k 1.求此数列的通项2.当k为何值时此数列为等差数列

题目

已知数列的前n项和为sn,且sn=2n^2-3n=k 1.求此数列的通项2.当k为何值时此数列为等差数列
打错了 是sn=2n^2-3n+k

答案

1.
Sn=2n^2-3n+k
S(n+1)=2(n+1)^2-3(n+1)+k
a(n+1)=S(n+1)-Sn=4n-1
an=4n-5
2.
只有当a1也满足通项公式时,即a1=(4*1-5)=-1时,{an}才是等差数列!
当n=1时,S1=a1=2-3+k=-1
k=0