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解
当n=1时,a1=s1=3-1=2
当n≥2时,
an=sn-s(n-1)
=3n-n²-[3(n-1)-(n-1)²]
=3n-n²-(3n-3-n²+2n-1)
=3n-n²-(-n²+5n-4)
=-2n+4
当n=1时,a1=-2×1+4=2
∴an=-2n+4
当n=1时,a1=s1=3-1=2
当n≥2时,
an=sn-s(n-1)
=3n-n²-[3(n-1)-(n-1)²]
=3n-n²-(3n-3-n²+2n-1)
=3n-n²-(-n²+5n-4)
=-2n+4
当n=1时,a1=-2×1+4=2
∴an=-2n+4
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