Let k , p be positive integers. Is there a closed form for the sums ∑ i = 0 p ( k i ) ( k + p − i p − i ) , or ∑ i = 0 p ( k − 1 i ) ( k + p − i p − i ) ?

Question

Let   k  ,  p be positive integers. Is there a closed form for the sums      ∑          i      =      0              p                          (                    k      i                      )                                (                            k        +        p        −        i                    p        −        i                            )              , or      ∑          i      =      0              p                          (                            k        −        1            i                      )                                (                            k        +        p        −        i                    p        −        i                            )              ?

Nkgopotsev1g · Accepted Answer

Derived the following simple inequality (confirmed numerically): for any   0  &amp;lt;  s  &amp;lt;  1      /    2,      ∑          i      =      0        p                                            (                                    k          i                                      )                                                                    (                                                k            +            p            −            i                                p            −            i                                                )                                ≤      1          (      1      −      2      s              )        k                        (                  1      −      s        s                      )              p    .Given   0  &amp;lt;  s  &amp;lt;  1      /    2, let   X be a binomial  (  k  ,  s  ) random variable, and   Y a binomial  (  k  +  p  −  X  ,  t  ) random variable, where   t  =  s      /    (  1  −  s  )    (  ∈  (  0  ,  1  )  ). Then, by the law of total probability,      P    (  X  +  Y  =  p  )  =      ∑          i      =      0        p              P        (    X    +    Y    =    p          |        X    =    i    )          P        (    X    =    i    )    =      ∑          i      =      0        p              P        (    Y    =    p    −    i          |        X    =    i    )          P        (    X    =    i    )    .Noting that      P    (  X  =  i  )  =                              (                            k        i                              )                          s    i    (  1  −  s      )          k      −      i      and      P    (  Y  =  p  −  i      |    X  =  i  )  =                              (                                      k          +          p          −          i                          p          −          i                                      )                          t          p      −      i        (  1  −  t      )    k    ,and using      s          (      1      −      s      )      t        =  1  ,we get      P    (  X  +  Y  =  p  )  =  (  1  −  s      )    k    (  1  −  t      )    k        t    p        ∑          i      =      0        p                                            (                                    k          i                                      )                                                                    (                                                k            +            p            −            i                                p            −            i                                                )                                .Finally, from       P    (  X  +  Y  =  p  )  ≤  1 and  (  1  −  s      )    k    (  1  −  t      )    k        t    p    =  (  1  −  2  s      )          k                  (            s          1      −      s                          )              p    ,it follows that      ∑          i      =      0        p                                            (                                    k          i                                      )                                                                    (                                                k            +            p            −            i                                p            −            i                                                )                                ≤      1          (      1      −      2      s              )        k                        (                  1      −      s        s                      )              p    .

Let k,p be positive integers. Is there a closed form for the sums Sum_(i=0)^p (k i) (k+p-i p-i), or Sum_(i=0)^p (k-1 i) (k+p-i p-i)?

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