Сombinatorial identity

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I try to prove the following hypothesis $$\sum_{i=0}^k \frac{(-1)^i (n+i-1)(n+k-2)!}{i!(k-1)!(n-1)!}{{n+2k-i-1} \choose {2k}}=\frac {{n+k-1\choose k}{n-2+k\choose k}}{k+1}.$$ In the Maple:

gip:=proc (n,i) simplify(binomial(n+i-1,i)*binomial(n-1-1+i,i)/(i+1)) end proc;

l:=proc (n,k) simplify(sum('(-1)^i*(n+i-1)*(n+k-2)!*binomial(n+2*k-i-1,2*k)/(i!(k-i)!(n-1)!)','i' = 0 .. k)) end proc;

for j from 0 by 1 to 12 do print(l(3,j),gip(3,j),l(4,j),gip(4,j),l(5,j),gip(5,j),l(6,j),gip(6,j),l(7,j),gip(7,j)) end do;

1, 1, 1, 1, 1, 1, 1, 1, 1, 1

              3, 3, 6, 6, 10, 10, 15, 15, 21, 21


           6, 6, 20, 20, 50, 50, 105, 105, 196, 196


        10, 10, 50, 50, 175, 175, 490, 490, 1176, 1176


      15, 15, 105, 105, 490, 490, 1764, 1764, 5292, 5292


    21, 21, 196, 196, 1176, 1176, 5292, 5292, 19404, 19404


   28, 28, 336, 336, 2520, 2520, 13860, 13860, 60984, 60984


  36, 36, 540, 540, 4950, 4950, 32670, 32670, 169884, 169884


  45, 45, 825, 825, 9075, 9075, 70785, 70785, 429429, 429429

55, 55, 1210, 1210, 15730, 15730, 143143, 143143, 1002001, 1002001

66, 66, 1716, 1716, 26026, 26026, 273273, 273273, 2186184, 2186184

78, 78, 2366, 2366, 41405, 41405, 496860, 496860, 4504864, 4504864

91, 91, 3185, 3185, 63700, 63700, 866320, 866320, 8836464, 8836464

But $$\sum_{i=0}^k \frac{(-1)^i (n+i-1)(n+k-2)!}{i!(k-1)!(n-1)!}{{n+2k-i-1} \choose {2k}}=\sum_{i=0}^k \frac{(-1)^i (n+i-1)(n+k-2)!k}{i!k!(n-2)!(n-1)}{{n+2k-i-1} \choose {2k}}=\\={n-2+k\choose k} \sum_{i=0}^k \frac{(-1)^i (n+i-1)k}{i!(n-1)}{{n+2k-i-1} \choose {2k}}.$$

And I should prove that $\sum_{i=0}^k \frac{(-1)^i (n+i-1)k}{i!(n-1)}{{n+2k-i-1} \choose {2k}}=\frac {{n+k-1\choose k}}{k+1}.$

I tried use Mathematical induction method (about k) and Chu's Theorem. The second way was using formula ${r+k+1 \choose k}=\sum_{i=0}^k {r+i \choose i}.$ I cannot prove it. Maple prove that it is false:

l1:=proc (n,k) simplify(sum('(-1)^i*(n+i-1)kbinomial(n+2*k-i-1,2*k)/(i!*(n-1))','i' = 0 .. k)) end proc; gip1:=proc (n,i) simplify(binomial(n+i-1,i)/(i+1)) end proc;

for j from 0 by 1 to 12 do print(l1(3,j),gip1(3,j),l1(4,j),gip1(4,j),l1(5,j),gip1(5,j)) end do; 0, 1, 0, 1, 0, 1

                   3/2, 3/2, 2, 2, 5/2, 5/2


                  17, 2, 115/3, 10/3, 75, 5


              111/2, 5/2, 313/2, 5, 2975/8, 35/4


                130, 3, 1346/3, 7, 7679/6, 14


            505/2, 7/2, 6205/6, 28/3, 27595/8, 21


                435, 4, 2065, 12, 31579/4, 30


                                     385679
           1379/2, 9/2, 22351/6, 15, ------, 165/4
                                       24


              1028, 5, 18692/3, 55/3, 29949, 55


          2925/2, 11/2, 19659/2, 22, 416847/8, 143/2


                                    1029535
              2005, 6, 44405/3, 26, -------, 91
                                      12


        5335/2, 13/2, 128755/6, 91/3, 1080233/8, 455/4


              3462, 7, 30146, 35, 409327/2, 140

Where is my eror?

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In the definition of l(n,k), why do you have $(k-i)!$ in the denominator instead of $k!$?
There is a line in the Maple code that defines $l$ as a procedure. Towards the end of the line, one of the factors is $(k-i)!$, in the denominator. None of the factors in the MathJax formula above it contain $(k-i)!$.