closed form for $\sum_{k=1}^n\left[ 2^k\binom{2n-k}{n-k}-2^{k+1}\binom{2n-k-1}{n-k-1}\right ]k^s$

Question

$$\sum_{k=1}^n\left[ 2^k\binom{2n-k}{n-k}-2^{k+1}\binom{2n-k-1}{n-k-1}\right ]k^s$$

by some steps I got $$\sum_{k=1}^n\left[ 2^k\binom{2n-k}{n-k}-2^{k+1}\binom{2n-k-1}{n-k-1}\right ]k^s=\sum_{k=1}^n2^k\binom{2n-k-1}{n}\frac{k^{s+1}}{n-k}$$

and considred $$\sum_{k=1}^n\binom{2n-k-1}{n}(2x)^ky^{n-k-1}=f(x,y)$$

the sum is $$xD^s_{x=1}(\int_0^1 f(x,y)dy)$$

$xD^s_{x=1}$ mean $d/dx$ then multiply by x then $d/dx$ then multiply ... s+1 time

but I have no idea how to continue :)

can I continue by this way ? or there is better way ?

if there is no closed form how to prove for $s=1,3,3,4$ ??

Answer 1

I do not know how much this could be of any interest to you, so forgive me if I am off-topic.

Let us consider, for $s>0$, $$A_s=\sum_{k=1}^n\left[ 2^k\binom{2n-k}{n-k}-2^{k+1}\binom{2n-k-1}{n-k-1}\right ]k^s$$ and $$B_s=\frac{1}{2}\frac{A_s}{\binom{2 n-1}{n-1}-2 \binom{2 n-2}{n-2}}$$ Then $$B_1=\, _3F_2(2,2,1-n;1,2-2 n;2)$$ $$B_2=\, _4F_3(2,2,2,1-n;1,1,2-2 n;2)$$ $$B_3=\, _5F_4(2,2,2,2,1-n;1,1,1,2-2 n;2)$$ $$B_4=\, _6F_5(2,2,2,2,2,1-n;1,1,1,1,2-2 n;2)$$ $$B_5=\, _7F_6(2,2,2,2,2,2,1-n;1,1,1,1,1,2-2 n;2)$$ $$B_6=\, _8F_7(2,2,2,2,2,2,2,1-n;1,1,1,1,1,1,2-2 n;2)$$ This shows clearly the pattern.