设数列{An}满足a1=1.Sn=a1+a2+a3+…+an=n2 (1)当n>=2时.求Sn-Sn-1
问题描述:
设数列{An}满足a1=1.Sn=a1+a2+a3+…+an=n2 (1)当n>=2时.求Sn-Sn-1
(2)求数列的通项公式An
答
1.
Sn=A1+A2+A3+……+An=n^2
S(n-1)=A1+A2+A3+……+A(n-1)=(n-1)^2
Sn-S(n-1)=n^2-(n-1)^2=2n-1
2.
n>=2时
An=Sn-S(n-1)=2n-1
经验算A1=1也满足上式