binomial identity seemingly illogical and impossible. Is there any way it could be true? 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.

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).
