数列an满足a1=3/2,a(n+1)=an^2-an+1则m=1/a1+1/a2````1/a2009整数部分是

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