已知数列{an}的前n项和sn满足sn=an^2+bn,求证{an}是等差数列

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已知数列{an}的前n项和sn满足sn=an^2+bn,求证{an}是等差数列已知数列{an}的前n项和sn满足sn=an^2+bn,求证{an}是等差数列已知数列{an}的前n项和sn满足sn=an

已知数列{an}的前n项和sn满足sn=an^2+bn,求证{an}是等差数列
已知数列{an}的前n项和sn满足sn=an^2+bn,求证{an}是等差数列

已知数列{an}的前n项和sn满足sn=an^2+bn,求证{an}是等差数列
n=1时,a1=S1=a+b
n≥2时,Sn=a×n²+bn
S(n-1)=a×(n-1)²+b
两式相减得：an=Sn-S(n-1)=2a×n-a
∴a(n-1)=2a×(n-1)-a
∴an-a(n-1)=2a
∴﹛an﹜是等差数列
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