若数列{an}满足a1=1,且1/[a(n+1)]-1/an=1,则a1*a2+a2*a3+…+a2010*a2011=?

a(n+1)=2an+1即
a(n+1)+1=2(an+1)=2^n(a1+1)=2^(n+1)
所以
a(n+1)=2^(n+1)-1
an=2^n-1
a1/a2+a2/a3+…+an/a(n+1)
=1/3+3/7+...+(2^n-1)/[2^(n+1)-1]
=1/2+1/2+...+1/2
=n/2
a1/a2+a2/a3+…+an/a(n+1)
=1/3+3/7+...+(2^n-1)/[2^(n+1)-1]
=(1.5-0.5)/3+(3.5-0.5)/7+...+[(2^n-0.5)+0.5]/[2^(n+1)-1]
=n/2-0.5{1/3+1/7+...+1/[2^(n+1)-1]}
>n/2-0.5{1/3+1/6+...+1/[2^(n+1)-2^(n-1)]+1/[2^(n+1)-2^(n-1)]}

1/[a(n+1)]-1/an=1
1/a1=1
所以1/an是以1为首项1为公差的等差数列
1/an=1+1(n-1)=n
an=1/n

a1*a2+a2*a3+…+a2010*a2011
=(1/1)(1/2)+(1/2)(1/3)+.+(1/2010)(1/2011)
=1/(1*2)+1/(2*3)+.+1/(2010*2011)
=1-1/2+1/2-1/3+.+1/2010-1/2011
=1-1/2011
=2010/2011

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