数列an满足a1=1,an=1/2a(n-1)+1(n≥2) (1)若bn=an-2,求证bn为等比数列 (2)求an的通项公式
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![数列an满足a1=1,an=1/2a(n-1)+1(n≥2) (1)若bn=an-2,求证bn为等比数列 (2)求an的通项公式](/uploads/image/z/11711654-62-4.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2Can%3D1%2F2a%28n-1%29%2B1%28n%E2%89%A52%EF%BC%89+%EF%BC%881%EF%BC%89%E8%8B%A5bn%3Dan-2%2C%E6%B1%82%E8%AF%81bn%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97+%EF%BC%882%EF%BC%89%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
数列an满足a1=1,an=1/2a(n-1)+1(n≥2) (1)若bn=an-2,求证bn为等比数列 (2)求an的通项公式
数列an满足a1=1,an=1/2a(n-1)+1(n≥2) (1)若bn=an-2,求证bn为等比数列 (2)求an的通项公式
数列an满足a1=1,an=1/2a(n-1)+1(n≥2) (1)若bn=an-2,求证bn为等比数列 (2)求an的通项公式
1,an=1/2a(n-1)+1,an-2=1/2[a(n-1)-2],bn=1/2b(n-1),所以bn为等比为2的等比数列 ;
2,b1=a1-2=1-2=-1,
所以bn=-1(1/2)^(n-1)=-(1/2)^(n-1),所以an=bn+2=2-(1/2)^(n-1).
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