已知数列an满足a1=1,a(n+1)=Sn+(n+1)(n属于自然数),证明数列{an+1}是等比数列.
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![已知数列an满足a1=1,a(n+1)=Sn+(n+1)(n属于自然数),证明数列{an+1}是等比数列.](/uploads/image/z/10226166-6-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2Ca%28n%2B1%29%3DSn%2B%28n%2B1%29%28n%E5%B1%9E%E4%BA%8E%E8%87%AA%E7%84%B6%E6%95%B0%29%2C%E8%AF%81%E6%98%8E%E6%95%B0%E5%88%97%EF%BD%9Ban%2B1%EF%BD%9D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97.)
已知数列an满足a1=1,a(n+1)=Sn+(n+1)(n属于自然数),证明数列{an+1}是等比数列.
已知数列an满足a1=1,a(n+1)=Sn+(n+1)(n属于自然数),
证明数列{an+1}是等比数列.
已知数列an满足a1=1,a(n+1)=Sn+(n+1)(n属于自然数),证明数列{an+1}是等比数列.
a(n+1)=Sn+n+1
a(n)=Sn-a(n)+n-1+1
化简可得
(a(n+1)+1)/(a(n)+1)=2
所以{an+1}是等比数列
a(n+1)=Sn+(n+1)
an=Sn-1 +n
两式做差
得a(n+1)=2an+1
两边都加1
得到等比数列,公比为2
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