Verify this sum $\sum_{j=0}^{2n}\frac{{2j \choose j}{2n \choose j}}{(-2)^j(2j-1)}=-(4n+1)\frac{(2n-1)!!}{(2n)!!}$

by coffeee   Last Updated May 16, 2019 03:20 AM

How to show that

$$\sum_{j=0}^{2n}\frac{{2j \choose j}{2n \choose j}}{(-2)^j(2j-1)}=-(4n+1)\frac{(2n-1)!!}{(2n)!!}$$

This double ${2j \choose j}{2n \choose j}$ is given me a difficult to to simplify the sum.



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