已知数列{an}的前n项和Sn=-2n2+3n,则数列{an}的通项公式为______

已知数列{an}的前n项和Sn=-2n2+3n,则数列{an}的通项公式为______.
2025-12-17 21:07:46
推荐回答(1个)
回答1:

∵Sn=-2n2+3n,
∴a1=S1=-2+3=1,
an=Sn-Sn-1=(-2n2+3n)-[-2(n-1)2+3(n-1)]
=5-4n.
当n=1时,5-4n=1=a1
∴an=5-4n,
故答案为:an=5-4n.