Combinations series: $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$ 
Evaluate $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$ for $n=10$.

Attempt: I'll deal with the case n being even, as we need to evaluate for n=10.
the numerator is 
$${n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots$$
$$=\sum_{r=odd} {n \choose r}(n-r)^3$$(not sure if this is a correct notation).
$$=\sum_{r=odd}{n \choose n-r}r^3=\sum_{r=odd} {n \choose r}r^3$$(parity being same as n is even and r is odd, although I don't think this matters much).
Using the identity ${n \choose r}=\frac{n}{r} {n-1 \choose r-1}$ repeatedly in following steps,
$$=n\sum_{r=even} {n-1 \choose r-1}r^2$$
$$=[n(n-1)](1+\sum_{r=odd} {n-2 \choose r-2}[(r-2+3)+\frac{1}{r-1}]$$
$$=[n(n-1)](1+(n-2)\sum_{r=even}{n-3 \choose r-3}+3\sum_{r=odd}{n-2 \choose r-2}+\frac{1}{n-1} \sum_{r=even}{n-1 \choose r-1} -1)$$
$$=[n(n-1)]((n-2)\cdot 2^{n-4} +3\cdot 2^{n-4}+\frac{2^{n-2}}{n-1}$$
This simplfies to $n \cdot 2^{n-4} (n^2+7n-4)$.
Which is incorrect. The answer for $n=10$ (numerator/denominator is given as $\frac{1}{16}$).
Where am I going wrong?
Also the hint given for this problem was "expand $\frac{(e^x+1)^n - (e^x-1)^n}{2}$ in two different ways". I didn't quite understand this approach?
Could someone please explain this approach and any other approach also?

Here's an alternative approach that does not depend on the hint.
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=\binom{n}{k+1}(k+1)^3$ to obtain
\begin{align}
&\sum_{k\ge 0} \binom{n}{2k+1}(2k+1)^3 \\
&= \sum_{k\ge 0} \binom{n}{k+1}(k+1)^3 \frac{1+(-1)^k}{2} \\
&= \sum_{k\ge 0} \frac{n}{k+1}\binom{n-1}{k}(k+1)^3 \frac{1+(-1)^k}{2} \\
&= n\sum_{k\ge 0} \binom{n-1}{k}(k+1)^2 \frac{1+(-1)^k}{2} \\
&= n\sum_{k\ge 0} \binom{n-1}{k}\left(2\binom{k}{2}+3k+1\right) \frac{1+(-1)^k}{2} \\
&= n\sum_{k\ge 0} \left(2\binom{n-1}{2}\binom{n-3}{k-2}+3(n-1)\binom{n-2}{k-1}+\binom{n-1}{k}\right) \frac{1+(-1)^k}{2} \\
&= \frac{n}{2}\left(2\binom{n-1}{2}\sum_{k\ge 0}\binom{n-3}{k-2}+3(n-1)\sum_{k\ge 0}\binom{n-2}{k-1}+\sum_{k\ge 0}\binom{n-1}{k}\right) \\
&+ \frac{n}{2}\left(2\binom{n-1}{2}\sum_{k\ge 0} \binom{n-3}{k-2}(-1)^k+3(n-1)\sum_{k\ge 0} \binom{n-2}{k-1}(-1)^k+\sum_{k\ge 0} \binom{n-1}{k}(-1)^k\right)  \\
&= \frac{n}{2}\left(2\binom{n-1}{2}2^{n-3}+3(n-1)2^{n-2}+2^{n-1}\right) \\
&+ \frac{n}{2}\left(2\binom{n-1}{2}(1-1)^{n-3}+3(n-1)(1-1)^{n-2}+(1-1)^{n-1}\right)  \\
&= 2^{n-4} n \left(2\binom{n-1}{2}+6(n-1)+4\right) \\
&+ \frac{n}{2}\left(2\binom{n-1}{2}[n=3]+3(n-1)[n=2]+[n=1]\right)  \\
&= 2^{n-4} n^2 (n+3) + 3[n=3]+3[n=2]+\frac{1}{2}[n=1]
\end{align}
So the fraction for even $n \ge 4$ is
$$\frac{2^{n-4} n^2 (n+3)}{2^n n^2 (n+3)} = \frac{1}{16}.$$
