Question

已知数列{an}对于任意的p、q∈N*,满足ap+q=ap+aq且a2=2,则1...

已知数列{an}对于任意的p、q∈N*,满足ap+q=ap+aq且a2=2,则1a1a2+1a2a3+…+1a2008a2009=______.

Answer 1

数列{an}对于任意的p、q∈N*,满足ap+q=ap+aq且a2=2,所以a2=a1+a1且a1=1,
所以an+1=an+1,数列是等差数列,an=n,所以
1
a1a2
+
1
a2a3
+…+
1
a2008a2009
=-(
1
a2
-
1
a1
+
1
a3
-
1
a2
+…+
1
a2009
-
1
a2008
)=1-
1
2009
=
2008
2009
.
故答案为:
2008
2009
.