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?
 A: Following the given hint, we have that
$$\begin{align}\sum_{r \text{ odd}} {10 \choose r}(10-r)^3&=\frac{1}{2}\left[\left((e^x+1)^{10}-(e^x-1)^{10}\right)'''\right]_{x=0}\\
&=\left[360(e^x+1)^7e^{3x}+135(e^x+1)^8e^{2x}+5(e^x+1)^9e^x\right.\\
&\quad \left.-360(e^x-1)^7e^{3x}-135(e^x-1)^8e^{2x}-5(e^x-1)^9e^x\right]_{x=0}\\
&=360\cdot 2^7+135\cdot 2^8+5\cdot 2^9.
\end{align}.$$
The same approach works for any integer $n\geq 4$:
$$\begin{align}\sum_{r \text{ odd}} {n \choose r}(n-r)^3
&=\frac{1}{2}\left[\left((e^x+1)^{n}-(e^x-1)^{n}\right)'''\right]_{x=0}\\
&=3\binom{n}{3}2^{n-3}+3\binom{n}{2}2^{n-2}+n2^{n-2}
=\frac{n^2(n+3)2^n}{16}.
\end{align}$$
P.S. We could also use the Taylor series of $e^x$: for $n\geq 4$
$$\begin{align}\frac{1}{2}\left[\left((e^x+1)^{n}-(e^x-1)^{n}\right)'''\right]_{x=0}
&=\frac{3!}{2}[x^3]\left((e^x+1)^{n}-(e^x-1)^{n}\right)\\
&=3[x^3]\left(2+x+\frac{x^2}{2}+\frac{x^3}{6}\right)^{n}\\
&=3[x^3]n2^{n-1}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)\\
&\quad  +3[x^3]\binom{n}{2}2^{n-2}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^2\\
&\quad  +3[x^3]\binom{n}{3}2^{n-3}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^3\\
&=n2^{n-2}+3\binom{n}{2}2^{n-2}+3\binom{n}{3}2^{n-3}\\
&=\frac{n^2(n+3)2^n}{16}.
\end{align}$$
A: 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
$$\frac{2^{n-4} n^2 (n+3)}{2^n n^2 (n+3)} = \frac{1}{16}.$$
A: A quick note on notation: you can use the notation I've used in this answer, or alternatively just use regular notation and $2r$ / $2r+1$ to denote even and odd, such as
$$
\sum_{r \in \mathbb Z} \binom{n}{2r+1} (n-(2r+1))^3
$$
Your derivation for even $n$ seems pretty much correct; the only issue I can see is that
$$
\sum_{\text{$r$ odd}} \binom{n-2}{r-2} = 2^{n-3},
$$
and you seem to have incorrectly used $2^{n-4}$; carrying through this correction yields the correct answer of $\frac 1 {16}$ for the $n = 10$ case. Your formula $\binom{n}{r} = \frac n r \binom{n-1}{r-1}$ of course breaks down when $r = 0$, but this doesn't actually end up being an issue here because those terms reduce to $0$ anyway (in general make sure you pay attention to this possibility though).
Of course this will yield a solution if you do it for $n$ odd as well, but the solution suggested by the hint is perhaps cleaner. With regards to that, here's a hint hint: there are two natural ways to expand the expression they give, one of which uses the formula
$$
x^n - y^n = (x - y) \left(x^{n-1} + x^{n-2} y + \ldots + y^{n-1}\right),
$$
and the other of which uses the binomial expansion on each of the two terms $(e^x + 1)^n$ and $(e^x - 1)^n$. This second will yield something that can quite clearly be transformed into the expression you want to evaluate; transforming the first as well should then give you a solution.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{{\large r = 1} \atop {\large r\ \mrm{odd}}}^{n}{n \choose r}
\pars{n - r}^{3} & =
\sum_{r = 0}^{n}{n \choose 2r + 1}
\bracks{\pars{n - \pars{2r + 1}}}^{\, 3}
\\[5mm] & =
\sum_{r = 0}^{n}{n \choose r}
\pars{n - r}^{\, 3}\,{1 - \pars{-1}^{r} \over 2}
\\[5mm] & =
\sum_{r = 0}^{n}{n \choose r}r^{\, 3}\,{1 - \pars{-1}^{\pars{n - r}}
\over 2}
\\[5mm] & =
\sum_{r = 0}^{n}{n \choose r}{1 - \pars{-1}^{\pars{n - r}} \over 2}
\bracks{z^{3}}3!\expo{rz}
\\[5mm] & =
3\bracks{z^{3}}\sum_{r = 0}^{n}{n \choose r}\pars{\expo{z}}^{r} -
3\pars{-1}^{n}\bracks{z^{3}}\sum_{r = 0}^{n}{n \choose r}
\pars{-\expo{z}}^{r}
\\[5mm] & =
3\bracks{z^{3}}\pars{1 + \expo{z}}^{n} -
3\pars{-1}^{n}\bracks{z^{3}}\pars{1 - \expo{z}}^{n}
\\[5mm] & =
\color{red}{\large 1 \over 16}\,2^{n}\,n^{2}\pars{n + 3} -
3\pars{-1}^{n}\bracks{z^{3}}\pars{1 - \expo{z}}^{n}\bracks{n \leq 3}
\end{align}

\begin{align}
&\bbox[10px,#ffd]{3\bracks{z^{3}}\pars{1 + \expo{z}}^{n}} =
3 \times 2^{n}\bracks{z^{3}}\pars{1 + {\expo{z} - 1 \over 2}}^{n}
\\[5mm] = &\
3 \times 2^{n}\bracks{z^{3}}\pars{1 + {1 \over 2}\,z +
{1 \over 4}\,z^{2} + {1 \over 12}\,z^{3}}^{n}
\\[5mm] = &\
3 \times 2^{n}\bracks{{1 \over 12}\,n + {n\pars{n - 1} \over 2}\,
{1 \over 4} + {n\pars{n - 1}\pars{n - 2} \over 6}\,{1 \over 8}} =
\color{red}{\large 1 \over 16}\,2^{n}\,n^{2}\pars{n + 3}
\end{align}
