Question

数列an满足a1=3/2,a(n+1)=an^2-an+1，则m=1/a1+1/a2+…+1/a

015

Answer 1

a（n+1）=an^2-an+1
a(n+1)-1 =an(an -1)
两边取倒数
1/[a（n+1）-1]=1/[an(an -1)]=1/(an -1) -1/an
就是
1/an =1/(an -1) -1/[a（n+1）-1]
所以
1/a1 = 1/(a1 -1) -1/(a2-1)
1/a2 =1/(a2-1) -1/(a3-1)
…
1/a2009 =1/(a2009-1) -1/(a2010-1)
1/a2010 =1/(a2010-1) -1/(a2011-1)

累加
m =1/(a1-1) -1/(a2011 -1)=2 -1/(a2011 -1)

又因为
a（n+1）=an^2-an+1=（an -1/2)^+3/4
==>a2 =7/4
a3 =17/8 >2
数列递增
....==>a2011 >2
1/(a2011 -1) <1
==> 1<m<2
则m的整数部分是1