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I am doing a project and somehow the alternating sum $\sum_{i=0}^{n}(-1)^{i}{{n+i}\choose{i}}$ comes up. I am not not sure if there is any use of this sum, but just think it is an interesting sum since it looks simple and yet I could not find a way to simply this sum (a surprise to me). I am wondering if anyone has experience with this sum and know how to simply it.

I know that $\sum_{i=0}^{n} {{n+i}\choose{i}} = {{2n+1}\choose{n}}$. But now the factor $(-1)^i$ causes a problem.

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    $\begingroup$ Duplicate of your previous post: math.stackexchange.com/questions/5110562/…. Please improve your original post instead of posting a new one. $\endgroup$ Commented Nov 24 at 8:09
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    $\begingroup$ By Mathematica: $\sum _{i=0}^n (-1)^i \binom{n+i}{i} = (-1)^n \binom{2 n+1}{n+1} \, _2F_1(1,2 n+2;n+2;-1)+2^{-n-1}$ $\endgroup$ Commented Nov 24 at 8:21
  • $\begingroup$ Many thanks for the kind reply. Now I learn something nontrivial and know that the sum is not as easy as it looks. $\endgroup$ Commented Nov 24 at 8:28
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    $\begingroup$ From $n=0$ to $n=9$, the values are alternately positive and negative, and their moduli are this sequence. $\endgroup$ Commented Nov 24 at 8:50
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    $\begingroup$ This question is similar to: Closed -form expression for the sun $\sum_{i=0}^{n}(-1)^{i}{{n+i}\choose{i}}$?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$ Commented Nov 24 at 10:15

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Reading the documentation of sequence $A026641$ is $OEIS$ (already mentioned by @J.G. in comments), and adapting it to your case, the generating function is $$\text{G.f.}=\frac{2}{-(1+4 x)+3\, \sqrt{1+4 x}}$$

Similarly, from the documentation, adapting R.J. Mathar's result, if

$$a_n=\sum_{i=0}^{n}(-1)^{i}{{n+i}\choose{i}}$$ then $$a_n=\frac{-(7n-4)\,a_{n-1}+2(2n-1)\,a_{n-2} }{2n}\quad \text{with}\quad a_0=1\quad \text{and}\quad a_1=-1$$

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  • $\begingroup$ Though it was not the task, at $n\gg1$ $$a_n=\sum_{i=0}^{n}(-1)^{i}{{n+i}\choose{i}}\sim\frac2{3\sqrt\pi}\,\frac{(-4)^n}{\sqrt n}$$ what already gives a decent accuracy at $n\sim1$ . For example, at $n=3$ the exact number is $-\,13$, and the asymptotics gives $-13.89802...$ - wolframalpha.com/… and wolframalpha.com/… $\endgroup$ Commented Nov 24 at 20:37
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    $\begingroup$ @Svyatoslav. You found the same as Vaclav Kotesovec, For once, I did not find anything in his blog. For $1.0$% relative error $n=18$; for $0.1$% relative error $n=181$; But any way thanks and cheers $\endgroup$ Commented Nov 25 at 3:44
  • $\begingroup$ More accurately, $$\sum_{i=0}^n(-1)^{i}{{n+i}\choose{i}}\sim\frac2{3\sqrt\pi}\,\frac{(-4)^n}{\sqrt n}\left(1-\frac{13}{72}\,\frac1{n}+O\Big(\frac1{n^2}\Big)\right)$$ what gives for $n=10\quad 122\, 464\,$ (exact) and $\,122\, 467,438...\,$ (approximation) $\endgroup$ Commented 2 days ago
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    $\begingroup$ @Svyatoslav. Beautiful would be an understatement ! Thanks and cheers, mon ami $\endgroup$ Commented 2 days ago

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