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From this Question:

How does one find the closed form for $(1)?$

$$\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose 2k}}{ak+b}=F(a,b)\tag1$$

$$\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose n-k}}{ak+b}=F(a,b)\tag2$$

This one is from the Question:

$F(0,1)=(-1)^n(2n+1)$

We got these two:

$F(1,1)=\frac{(-1)^n}{n}$

$F(2,1)=\frac{(-1)^n}{2n+1}$

Sibawayh
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    Do you have reason to expect that there is a closed form? – joriki Mar 27 '20 at 13:35
  • So far we got three particular results, so I suspect there should be a closed-form, but I doubt it is simple. – Sibawayh Mar 27 '20 at 13:44
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    Wolfram|Alpha mostly just returns hypergeometric series or "difference roots", but it gives interesting results for $F(2,3)$ and $F(2,5)$. – joriki Mar 27 '20 at 14:40
  • Your $F(1,1)$ seems to be incorrect, it should be $F(1,1)=\frac{(-1)^n(2n+1)-1}{2n(n+1)}$ (this also implies values for all $F(a,a)=\frac{1}{a}F(1,1)$) – Sil Mar 28 '20 at 12:32

1 Answers1

2

You can find a closed form as follow :

$$ F(a,b)=\left(\sum_{k=0}^n \int_0^1(-4t)^{ak+b-1}\binom{n+k}{2k}dt \right) $$

Finite sum

$$ F(a,b)=\left( \int_0^1\sum_{k=0}^n(-4t)^{ak+b-1}\binom{n+k}{n-k}dt \right) $$

Then using the coefficient extraction function $[X^i]$ for a given $i$.

\begin{align*} \sum_{k=0}^n(-4t)^{ak+b-1}\binom{n+k}{n-k} & = \sum_{k=0}^n(-4t)^{ak+b-1}[X^{n-k}](1+X)^{n+k} \\ & = \sum_{k=0}^n(-4t)^{ak+b-1}[X^n]z^k(1+X)^{n+k} \\ & = [X^n](1+X)^n(-t)^{a+b-1}\sum_{k=0}^n(-4^at^a)^{k}X^k(1+X)^{k} \\ & = [X^n](1+X)^n(-t)^{a+b-1}\dfrac{1-(-4^at^aX(1+X))^{n+1}}{1+4^at^aX(1+X)} \\ &= [X^n](1+X)^n(-t)^{a+b-1}\dfrac{1}{1+4^at^aX(1+X)} \\ &=[X^n](-t)^{a+b-1}\dfrac{(1+X)^{n-1}}{1+4^at^aX} \\ & = [X^n](-t)^{a+b-1}(1+X)^{n-1}\sum_{i=0}^\infty(-1)^i(4^at^aX)^i \\ & = (-t)^{a+b-1}\sum_{i=0}^\infty(-1)^i[X^{n-i}](1+X)^{n-1}(4^at^a)\\ & = (-t)^{a+b-1}\sum_{i=0}^\infty(-1)^i\binom{n-1}{i}(4^at^a)^i\\ & = (-t)^{a+b-1}(1-4^at^a)^{n-1}\\ \end{align*}

From here the exercise can be ended.

Credits

This answer is inspired from the work in the answer here : sum of binomial series with alternate terms.

EDX
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