已知数列(an)的前n项之和为Sn,(1)Sn=-n²+2n,求通项公式

网友回答

已知数列(an)的前n项之和为Sn,(1)Sn=-n²+2n
n=1时,a1=S1=-1²+2=1
n>1时,an=Sn-S(n-1)
=[-n²+2n]-[-(n-1)²+2(n-1)]
=[-n²+2n]-[-n²+2n-1+2n-2]
=-n²+2n+n²-2n+1-2n+2
=-2n+3
而an=-2n+3满足n=1的情况
所以,综上所述
an=-2n+3