以知等差数列an前N项和为Sn=-2n^2-n,求通项an的表达式为什么S(n-1)=-2(n-1)²-(n-1)=-2n²+3n-1呢?
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![以知等差数列an前N项和为Sn=-2n^2-n,求通项an的表达式为什么S(n-1)=-2(n-1)²-(n-1)=-2n²+3n-1呢?](/uploads/image/z/3959374-22-4.jpg?t=%E4%BB%A5%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E5%89%8DN%E9%A1%B9%E5%92%8C%E4%B8%BASn%3D-2n%5E2-n%2C%E6%B1%82%E9%80%9A%E9%A1%B9an%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%E4%B8%BA%E4%BB%80%E4%B9%88S%28n-1%29%3D-2%28n-1%29%26sup2%3B-%28n-1%29%3D-2n%26sup2%3B%2B3n-1%E5%91%A2%EF%BC%9F)
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以知等差数列an前N项和为Sn=-2n^2-n,求通项an的表达式为什么S(n-1)=-2(n-1)²-(n-1)=-2n²+3n-1呢?
以知等差数列an前N项和为Sn=-2n^2-n,求通项an的表达式
为什么S(n-1)=-2(n-1)²-(n-1)=-2n²+3n-1呢?
以知等差数列an前N项和为Sn=-2n^2-n,求通项an的表达式为什么S(n-1)=-2(n-1)²-(n-1)=-2n²+3n-1呢?
n>=2
S(n-1)=-2(n-1)²-(n-1)=-2n²+3n-1
所以an=Sn-S(n-1)=-4n+1
a1=S1=-2-1=-3
符合n>=2时的an=-4n+1
所以an=-4n+1
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