Coolefficents

Probability Level 5

There exists a sequence $\left\{a_n\right\}$ such that the equation below holds true for every non-negative integers $n$ . $\sum_{k=0}^{n}\binom nka_k=n^{2018}$ Find $\ln(a_{2018})$ .

The answer is 13343.4.

1 solution

Mark Hennings
Jun 16, 2018

Using the Binomial Transform, we obtain $a_n \; = \; \sum_{k=0}^n \binom{n}{k}(-1)^{n-k}k^{2018} \; = \; 2018! \left\{ {2018 \atop n } \right\}$ where $\left\{{N \atop n}\right\}$ denotes the Stirling number of the second kind. Thus $a_{2018} = 2018!$ , and so $\ln\,a_{2018} = \boxed{13343.4}$ .

Coolefficents

The answer is 13343.4.

1 solution

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