Find $\frac{({n\choose 1})(n-1)^3+({n\choose 3})(n-3)^3+...}{n^2(n+3)2^n}$ Find $\frac{{n\choose 1}(n-1)^3+{n\choose 3}(n-3)^3+...}{n^2(n+3)2^n}$
My attempt at this question was to first write the numerator as
$\sum_{k=1}^{n} {n\choose 2k-1}(n-(2k-1))^3$ and then expanding this
we would get expression like
$\sum_{k=1}^{n} {n\choose 2k-1}(2k-1)^3$ which if I write as
$\frac{1}{2}\sum_{r=0}^{n}{n\choose r}r^3$ I might be able to solve this using the repeated differentiation of $(1+x)^n$ ,
But overall I am not sure if my approach is correct the sum seemed to get more complicated , is there any more elegant ways to solve this ?
 A: You have $n$ people, of these you want to select an odd number of people to be in a committee, then from the non committee you choose a technical, marketing, and sales manager where one person can have more than one title. The number of possibilities are:
$$
\sum_{i\in\mathbb{N}}{\binom{n}{2i-1}\left(n-(2i-1)\right)^{3}}
$$
Which is the numerator of your equation. Now let’s count in a different way. We choose the managers first then select an odd number of people from the non managers. The following are the sum of possibilities from the case when all managers are different, one person hold two titles, and one person hold three titles:
$$
3!\cdot\binom{n}{3}\cdot 2^{n-4}+3\cdot2\cdot\binom{n}{2}\cdot 2^{n-3}+\binom{n}{1}\cdot 2^{n-2}
$$
Here I use the well known fact that there are $2^{m-1}$ odd - subset of a set with $m$ members. The two expressions are equal since both count the same possibilities, only with different method
A: Algebraic brute force way:
$$ S = \sum_{k=1}^n \binom{n}{2k+1} \left[ n- (2k+1) \right]^3= \sum_k \binom{n}{2k+1} \left[ \sum_j \binom{3}{j} (2k+1)^j n^{3-j}\right]= \sum_j \binom{3}{j} n^{3-j} \sum_k \binom{n}{2k+1} (2k+1)^j \tag{1}$$
We need to evaluate: $$\sum_k \binom{n}{2k+1} (2k+1)^j$$
Notice that $$x \frac{d}{dx} (1+x)^n= x\frac{d}{dx} \sum_k \binom{n}{k} x^k = \sum_k  \binom{n}{k}k x^k$$
Call $\phi= x \frac{d}{dx}$ then:
$$ \phi^j \left[ (1+x)^n\right] = \sum_k \binom{n}{k} k^j x^k $$
Now, we just need to take odd part of both sides and evaluate at one, which is given as:
$$ \phi^j \left[ (1+x)^n - (1-x)^n \right]|_{x=1} = \sum_k \binom{n}{2k+1} (2k+1)^j $$
Hence (1) becomes:
$$ S= \sum_j \binom{3}{j} n^{3-j} \phi^j \left[ (1+x)^n - (1-x)^n \right]_{x=1}$$
It maybe noted that $\phi^j$ can be expanded into a linear combination of ordinary derivatives weighted by Stirling numbers. Refer

