数列{an}满足a1=3/2,an+1=an^2-an+1(n属于正整数),则m=1/a1+1/a2+……+1/a2009的整数部分是?

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数列{an}满足a1=3/2,an+1=an^2-an+1(n属于正整数),则m=1/a1+1/a2+……+1/a2009的整数部分是?
数列{an}满足a1=3/2,an+1=an^2-an+1(n属于正整数),则m=1/a1+1/a2+……+1/a2009的整数部分是?

数列{an}满足a1=3/2,an+1=an^2-an+1(n属于正整数),则m=1/a1+1/a2+……+1/a2009的整数部分是?
由题知,a(n+1)-1=a(n)*(a(n)-1),1/(a(n+1)-1)=1/[a(n)*(a(n)-1)=1/(a(n)-1)-1/a(n);
得1/(a(n)-1)-1/(a(n+1)-1)=1/a(n),通过累加的方法得,
1/a1+1/a2+……+1/a2009= 1/(a1-1)-1/(a2010-1)=2-1/(a2010-1)
由a(n+1) - a(n)=(a(n)-1)^2≥0 ,即a(n+1)≥a(n),由a1=3/2,得a2=7/4,得a3=2又5/16.
所以,a2010≥a009≥a2008≥……≥a3>2,即 0

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