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There is binomial expression(s) written as $$\sum_{n\geqslant0}\frac{(-3n+2k-3)n!^2}{2(2n+1)(k-1)!^2(n-k+1)!^2 \binom{2n}{n}}=\begin{cases} 0 & \text{if $k=0$,} \\ -1 & \text{if $k\geqslant1$,} \end{cases}$$

which simplifies to $$\sum_{n\geqslant0}\frac{(3n-2k+1)\binom{n}{k}^2}{(2n+1) \binom{2n}{n}}=2\quad\text{for all $k\geqslant0$.}$$ Logically this looks like a physical impossibility. Would anyone believe that there is any way this could be true and if so how would one go about proving this paradox or seemingly impossibility.

John Bentin
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user158293
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  • omg yes i see i made a typo and the sum should be from 0 ! i will see if there is any possible way to edit the question. – user158293 Oct 08 '21 at 10:56
  • actually it should have been in both eq's. >-1 rather than >0 . It is just confusing whenever they use that for example .ge. rather than just a single inequality so i copied it wrong. – user158293 Oct 08 '21 at 11:02
  • book A=B p128(7.3.2) no need to go to chat i think this last ans. should be all he is asking – user158293 Oct 08 '21 at 12:33
  • if u want evidence then google "book A=B" – user158293 Oct 08 '21 at 13:37
  • I have checked your formulas against the source. Thank you for sharing this spectacular result with us! – John Bentin Oct 08 '21 at 14:12
  • now u can see why i first mistook the $n\geqslant0}$ as u wrote it for $n>0$ as that other lower bar is very difficult to discern. Still think it's better to use $n>-1$ even though he did not write it that way. – user158293 Oct 08 '21 at 14:17
  • I take your point about visibility; the slanted underbar in Mathjax is a bit too close to the $>$ (or $<$) sign, and $\geq$ would be better here. I prefer $\sum_{n=0}^\infty$to the unnatural $\sum_{n>-1}$ . – John Bentin Oct 08 '21 at 15:24
  • yea given that one is going to place a lower and upper limit the $\sum_{n=0}^{\infty}$ is better but if given that u are going to only place a single essentially lower limit prefer the $\sum_{n>-1}$ style as long as assume takes only integer values. – user158293 Oct 09 '21 at 01:33

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it seems that he just let say k=k'+1 in the first eq multiplied by -2 and then dropped the primes on k to get the 2nd expression(7.3.2). Also the first case "if k=0" must also be mult by -2 just as for the 2nd case. If k goes from 0 then since k=k'+1 then k' starts from -1 There is also one more issue is that if we were to say k' goes from one less then the 2nd should read after replacing k' by k then it should read k=-1,0,1,2... but then could argue that because then $\binom{n}{-1}$ is equal to 0 then we could change it to k=0,1,2...but equivalently we also already knew that for the original k first case that it was 0 anyway so just and (in fact must leave that out since the expression in not equal 2) which corresponds to k'=-1 which we leave off which after dropping primes the range is just as written in (7.3.2).

user158293
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  • actually i probably should not have asked the question for now assuming there are in fact no ambiguities but my answer choices were all correct then it does not seem that difficult though it sure initially seem that way at first. In short let k=k'+1 multiply by -2 and drop the primes on the 2nd expression and then only need take k' for 0,1,2,.... rather than the more direct -1,0,1,2... because of the $\binom {n}{-1}$ is 0 anyway. Hope there are no more caveats. – user158293 Oct 08 '21 at 13:53