数列an满足a1=1,当n≥2时an²-(n+2)*an-1*an+2*n*an-1²=0 求通项公式
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![数列an满足a1=1,当n≥2时an²-(n+2)*an-1*an+2*n*an-1²=0 求通项公式](/uploads/image/z/241014-30-4.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2C%E5%BD%93n%E2%89%A52%E6%97%B6an%26%23178%3B-%28n%2B2%29%2Aan-1%2Aan%2B2%2An%2Aan-1%26%23178%3B%3D0+%E6%B1%82%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
数列an满足a1=1,当n≥2时an²-(n+2)*an-1*an+2*n*an-1²=0 求通项公式
数列an满足a1=1,当n≥2时an²-(n+2)*an-1*an+2*n*an-1²=0 求通项公式
数列an满足a1=1,当n≥2时an²-(n+2)*an-1*an+2*n*an-1²=0 求通项公式
an²-(n+2)*an-1*an+2*n*an-1²=0
因式分解得(an-2a(n-1))(an-na(n-1))
则an-2a(n-1)=0 或an-na(n-1)=0
即an=2a(n-1) 或an/a(n-1)=n
故an=2^(n-1) 或 an=n!
an²-(n+2)*(an-1)*an+2*n*(an-1)²=0
所以(an-2(an-1))(an-n(an-1))=0
所以an=2an-1或者an=nan-1
前者是等比数列,an=2^(n-1)
后者,an=n!
(其实这题有问题,因为可以有些项满足an=2an-1,另一些项满足an=nan-1,这样都能使得an²-(...
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an²-(n+2)*(an-1)*an+2*n*(an-1)²=0
所以(an-2(an-1))(an-n(an-1))=0
所以an=2an-1或者an=nan-1
前者是等比数列,an=2^(n-1)
后者,an=n!
(其实这题有问题,因为可以有些项满足an=2an-1,另一些项满足an=nan-1,这样都能使得an²-(n+2)*(an-1)*an+2*n*(an-1)²的值为0,通项求不出来,但是题意应该是所有项都只满足两者中的一个,答案如上)
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