Let $f_{n}(x) = \dfrac{1}{n!}x^n(1-x)^n. $ Prove that the t-th derivative of the function evaluated at $0$ is an integer.

Question

Let $n \in \mathbb N$ and $$f_{n}(x) = \dfrac{1}{n!}x^n(1-x)^n. $$

Prove that for $t = 0,1,2, \cdots, f^{(t)}_n(0) \in \mathbb Z$ where $t$ is the t-th derivative.

My solution:

$$f_{n}(x) = \dfrac{1}{n!}x^n(1-x)^n = \dfrac{1}{n!}x^n \sum_{k=0}^{n} \binom{n}{k} (1)^{n-k}(-x)^k = \dfrac{1}{n!}\sum_{k=0}^{n}(-1)^k \binom{n}{k} x^{n+k}.$$

This the easy part. If $ t > n+k \quad\forall\, k \,\,\, ,f^{(t)}_{n}(x) = 0 \quad\forall\,x. $

I'm not able to understand how the proof will work when $t \leq n+k.$

Answer 1

Note that actually only the term with $n+k=t$ contributes to $f_n^{(t)}(0)$. Hence $$f_n^{(t)}(0) = \frac{1}{n!}(-1)^k{n \choose k}(n+k)!.$$ As ${n \choose k}$ and $\frac{(n+k)!}{n!}$ are integers, this term is always an integer.