已知数列前n项和为Sn,且Sn=-2n+3,求证:数列为等比数列
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已知数列前n项和为Sn,且Sn=-2n+3,求证:数列为等比数列
![已知数列前n项和为Sn,且Sn=-2n+3,求证:数列为等比数列](/uploads/image/z/19362682-10-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94Sn%3D-2n%2B3%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97)
证明:
因为数列前n项和为Sn,且Sn=-2n+3
所以:sn-1 = -2(n-1) + 3
= -2n + 2 + 3
= -2n + 5
所以:an = sn - sn-1
=-2n + 3 - ( -2n + 5 )
= -2n + 3 + 2n - 5
= -2
所以数列{an}是常数列,则数列{an}是等比数列
首项为a1 = -2 q = -2/-2 = 1
因为数列前n项和为Sn,且Sn=-2n+3
所以:sn-1 = -2(n-1) + 3
= -2n + 2 + 3
= -2n + 5
所以:an = sn - sn-1
=-2n + 3 - ( -2n + 5 )
= -2n + 3 + 2n - 5
= -2
所以数列{an}是常数列,则数列{an}是等比数列
首项为a1 = -2 q = -2/-2 = 1
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