当n=n-1时,Sn-1=a(n-1)²+b(n-1),Sn-Sn-1=an²+bn-a(n-1)²+b(n-1),即an(a的第n项)=an²+bn-an²-a+2an-bn+b=b-a+2an,即an=a-b,因为a和b是常数,且a-b是与n无关的常数,所以Sn=an²+bn是等差数列.
高中数学有关数列的证明题求证:数列{an}的前n项和Sn=an²+bn(a,b为常数)的充要条件是数列{an}
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