Question

数列an的前n项和为Sn=n^2+3n-1,则它的通项公式an=

急求具体具体过程，能详细就详细
额 打错了 是 Sn=n^2+3n+1

Answer 1

解
当n=1时：a1=1+3+1=4
当n≥2时：
an=Sn-S(n-1)
=n²+3n+1-(n-1)²-3(n-1)-1
=n²+3n+1-n²+2n-1-3n+3-1
=2n+2
当n=1时：a1=2×1+2=4
∴
an=2n+2

Answer 2

当n=1时sn=1+3+1=5
an
=Sn-Sn-1
=n^2+3n+1-[(n-1)^2+3(n-1)+1]
=n^2+3n+1-[(n^2-2n+1)+3n-3+1]
=n^2+3n+1-{n^2-2n+1+3n-3++1]
=n^2+3n+1-{n^2-n-1]
=n^2+3n+1-n^2+n+1
=4n+2
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