One of the problems from the 1990 Australian Mathematical Olympiad () asks one to prove the following result.
One solution in  makes use of the following identity:
Then setting , (1) becomes a telescoping sum:
Using a similar argument one can prove the more general identity
How about the sum ? Using (1) we find the recursion
Hence we have the recurrence relation
From this does not have a closed form solution but using induction on n we can prove the following relations found in .
Firstly note that when , the three expressions in (5) become , and , which are all equal to 1. Hence (6) holds for . Assume it holds for . That is,
Then using (5), on the one hand,
while on the other,
Equations (8) and (9) show that if (6) holds for , then it does for . By the principle of mathematical induction, (6) then is true for all integers .
Another related sum that can be expressed in terms of is . We have
Note that more identities with inverses of binomial coefficients are in  (and references therein), where the integral
 A. W. Plank, Mathematical Olympiads: The 1990 Australian Scene, University of Canberra, 1990.
 T. Mansour, Combinatorial Identities and Inverse Binomial Coefficients, available at http://math.haifa.ac.il/toufik/toupap02/qap0204.pdf