已知数列﹛an﹜的前n项和Sn=﹣4n²+3n,求证数列﹛an﹜是等差数列,并求其第11项

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/25 01:44:59

Sn=﹣4n²+3n
an=Sn-S(n-1)
=(-4n^2+3n)-[-4(n-1)^2+3(n-1)]
=-4n^2+3n+4(n-1)^2-3(n-1)
=-4n^2+3n+4n^2-8n+4-3n+3
=-8n+7
又因为
an-a(n-1)=-8n+7-[-8(n-1)+7]
=-8n+7+8n-8-7
=-8
所以数列﹛an﹜是等差数列
a11=-8*11+7=-81

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