IMO 1977 LL HUN25

Origin: HUN

Problem

Prove the identity (z + a)n = zn + a n  k=1 n k  (a −kb)k−1(z + kb)n−k.

Solution

Let fn(z) = zn + a n  k=1 n k  (a −kb)k−1(z + kb)n−k. We shall prove by induction on n that fn(z) = (z + a)n. This is trivial for n = 1. Suppose that the statement is true for some positive integer n −1. Then f ′ n(z) = nzn−1 + a n−1  k=1 n k  (n −k)(a −kb)k−1(z + kb)n−k−1 = nzn−1 + na n−1  k=1 n −1 k  (a −kb)k−1(z + kb)n−k−1 = nfn−1(z) = n(z + a)n−1. It remains to prove that fn(−a) = 0. For z = −a we have by the lemma of (SL81-13), fn(−a) = (−a)n + a n  k=1 n k  (−1)n−k(a −kb)n−1 = a n  k=0 n k  (−1)n−k(a −kb)n−1 = 0.