数列an满足a1=1,an=2an-1/(2+an-1) (n≥2),用数学归纳法求an的通项公式?
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![数列an满足a1=1,an=2an-1/(2+an-1) (n≥2),用数学归纳法求an的通项公式?](/uploads/image/z/1821665-65-5.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2Can%3D2an-1%2F%282%2Ban-1%29+%28n%E2%89%A52%29%2C%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%3F)
数列an满足a1=1,an=2an-1/(2+an-1) (n≥2),用数学归纳法求an的通项公式?
数列an满足a1=1,an=2an-1/(2+an-1) (n≥2),用数学归纳法求an的通项公式?
数列an满足a1=1,an=2an-1/(2+an-1) (n≥2),用数学归纳法求an的通项公式?
a2=2*1/(1+2)=2/3
类似的a3=1/2
a4=2/5
a5=1/3
注意,1=2/2,1/2=2/4,1/3=2/6
所以猜想an=2/(1+n)
①k=1已经成立
②假设k=n时an=2/(1+n)成立,k=n+1时,an+1=2an/(2+an)=2/(n+2)=2/((n+1)+1)
即对于k=n+1也成立
由一二可得,对任意n,an有通项公式an=2/(1+n)
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