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\sum {n \choose k} (a + k)^k (b - k)^{n-k} = \sum n! /t! (a + b)^t

Source: 6th QEDMO problem 4 (22. - 29. 8. 2009) https://artofproblemsolving.com/community/c1512515_qedmo_200507

May 8, 2021
SumalgebraCombinations

Problem Statement

Let aa and bb be two real numbers and let nn be a nonnegative integer. Then prove that ∑k=0n(nk)(a+k)k(b−k)n−k=∑k=0nn!t!(a+b)t\sum_{k=0}^{n} {n \choose k} (a + k)^k (b - k)^{n-k} = \sum_{k=0}^{n} \frac{n!}{t!} (a + b)^t
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