Question 11
(AUS 4) Prove that
$$\sum_{m=0}^{995}\frac{(-1)^m}{1991-m}\binom{1991-m}{m}=\frac1{1991}.$$
$$\sum_{m=0}^{995}\frac{(-1)^m}{1991-m}\binom{1991-m}{m}=\frac1{1991}.$$
IMO 1991 - Shortlisted problems Question 11
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