已知数列{an}的前n项和为Sn=n^2+2n,求数列{an}的通项公式

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2025-01-21 18:52:34
推荐回答(2个)
回答1:

先令n=1,求出a1=s1则n>=2时an=Sn-Sn-1再合并

回答2:

Sn=n^2+2n
S(n-1)=(n-1)^2+2(n-1)
=n^2-2n+1+2n-2
=n^2-1

an=Sn-S(n-1)
=n^2+2n-(n^2-1)
=2n+1