Is it an Algebra Problem or Calculus?

It is possible to write the sum in this way:

$S(n)= \sum_{i=0}^{\lfloor{\frac{n-1}{2}}\rfloor} {{n-1-i}\choose{i}}\frac{2^{n-1-2i}}{n-2i} = \frac{1}{2} \sum_{i=0}^{\lfloor{\frac{n-1}{2}}\rfloor} {{n-1-i}\choose{i}}\int_{0}^{2}x^{n-2i-1}dx=\frac{1}{2}\int_{0}^{2}dx \sum_{i=0}^{\lfloor{\frac{n-1}{2}}\rfloor} {{n-1-i}\choose{i}}x^{n-2i-1}$

Let $f(n)= \sum_{i=0}^{\lfloor{\frac{n-1}{2}}\rfloor} {{n-1-i}\choose{i}}x^{n-2i-1}$

Using Pascal's identity for binomial coefficient we have

$f(n+1)=\sum_{i=0}^{\lfloor{\frac{n}{2}}\rfloor} {{n-i}\choose{i}}x^{n-2i}=\sum_{i=0}^{\lfloor{\frac{n-1}{2}}\rfloor} {{n-1-i}\choose{i}}x^{n-2i}+\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor} {{n-1-i}\choose{i-1}}x^{n-2i}$

Setting $i=k-1$ in $f(n)$ and using the property of the floor function we could write the following linear recurrence relation:

$f(n+1)=xf(n)+f(n-1)$

with $f(1)=1$ and $f(2)=x$ .

Solving the recurrence relation in the standard way we have:

\[

f(n)= \frac{1}{2^{n}}\frac{1}{(x-\sqrt{x^2+4})^2+4} \left[ 8\left(x+\sqrt{x^2+4}\right)^{n-1}+2\left(x-\sqrt{x^2+4}\right)^{n+1}\right]

\]

Taking the integral we finally have

$S(n) = \frac{ \left(1+\sqrt{2}\right)^{n}+\left(1-\sqrt{2}\right)^{n}-1+(-1)^{n+1}}{2n}$

So:

$S(20)=\frac{5654884}{5}$