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
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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的通项公式
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*
,把a(n)+2a(n+1)+6n+4=0写出bn=an+2n这种形式
a(n)+2n=2[a(n+1)+2(n+1)]
{a(n)+2n}即{b(n)}是 首项为3 公比为1/2的等比数列
a(n)+2n=3*(1/2)^(n-1)
故a(n)=3*(1/2)^(n-1)-2n
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