已知数列an,a1=1,an+2a(n+1)+6n+4=0,若bn=an+2n,(1)求证bn是等比数列(2)求数列an
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已知数列an,a1=1,an+2a(n+1)+6n+4=0,若bn=an+2n,(1)求证bn是等比数列(2)求数列an的通项公式
![已知数列an,a1=1,an+2a(n+1)+6n+4=0,若bn=an+2n,(1)求证bn是等比数列(2)求数列an](/uploads/image/z/18095177-65-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%2Ca1%3D1%2Can%2B2a%EF%BC%88n%2B1%EF%BC%89%2B6n%2B4%3D0%2C%E8%8B%A5bn%3Dan%2B2n%2C%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81bn%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97an)
证明:1)b(n) = a(n) + 2*n ==> a(n) = b(n) - 2*n 且 a(n+1) = b(n+1) - 2*(n+1) 带入已知的递推式有 b(n) - 2*n + 2*(b(n+1) -2*(n+1)) + 6*n + 4 = 0,化简得 b(n) + 2*b(n+1) = 0 即b(n+1) = (-1/2)*b(n) 且 b(1) = a(1) + 2 = 3,所以b(n)是首项为3,公比为-1/2的等比数列
2)从1)中可得 b(n) = b(1)*(-1/2)^(n-1) = 3*(-1/2)^(n-1) ==> a(n)=b(n)-2*n = 3*(-1/2)^(n-1) - 2*n,n∈N*
2)从1)中可得 b(n) = b(1)*(-1/2)^(n-1) = 3*(-1/2)^(n-1) ==> a(n)=b(n)-2*n = 3*(-1/2)^(n-1) - 2*n,n∈N*
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