One of the problems from the 1990 Australian Mathematical Olympiad ([1]) asks one to prove the following result.

One solution in [1] 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 [2].

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 [2] (and references therein), where the integral

is utilised.

#### References

[1] A. W. Plank, Mathematical Olympiads: The 1990 Australian Scene, University of Canberra, 1990.

[2] T. Mansour, Combinatorial Identities and Inverse Binomial Coefficients, available at http://math.haifa.ac.il/toufik/toupap02/qap0204.pdf

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