<s>[INST] 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?
 [/INST] 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 $n\geq 4$
$$\begin{align}\frac{1}{2}\left[\left((e^x+1)^{n}-(e^x-1)^{n}\right)'''\right]_{x=0}
&=\frac{3!}{2}[x^3]\left((e^x+1)^{n}-(e^x-1)^{n}\right)\\
&=3[x^3]\left(2+x+\frac{x^2}{2}+\frac{x^3}{6}\right)^{n}\\
&=3[x^3]n2^{n-1}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)\\
&\quad  +3[x^3]\binom{n}{2}2^{n-2}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^2\\
&\quad  +3[x^3]\binom{n}{3}2^{n-3}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^3\\
&=n2^{n-2}+3\binom{n}{2}2^{n-2}+3\binom{n}{3}2^{n-3}\\
&=\frac{n^2(n+3)2^n}{16}.
\end{align}$$
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