Denote the sum S n := ∑ k = 0 n ( n k ) ( − 1 ) n − k n + k + 1

Question

Denote the sum      S    n    :=      ∑          k      =      0        n                                (                            n        k                              )                                (      −      1              )                  n          −          k                            n      +      k      +      1

Camron Herrera · Accepted Answer

We have                              ∫                      0                                1                          (        x        (        1        −        x        )                  )          n                         dx                            =                  ∫                      0                                1                          (        x        −                  x          2                          )          n                         dx                                          =                  ∫                      0                                1                                    ∑                      k            =            0                                n                                                              (                                            n            k                                              )                                                x          k                (        −                  x          2                          )                      n            −            k                                   dx                                          =                  ∑                      k            =            0                                n                                                              (                                            n            k                                              )                                      (        −        1                  )                      n            −            k                                    ∫                      0                                1                                    x                      2            n            −            k                                   dx                                          =                  ∑                      k            =            0                                n                                                              (                                            n            k                                              )                                                            (            −            1                          )                              n                −                k                                                          2            n            −            k            +            1                                                                      =          ⋆                          ∑                      k            =            0                                n                                                              (                                            n                          n              −              k                                                          )                                                            (            −            1                          )              k                                            2            n            −            (            n            −            k            )            +            1                                                            =                  ∑                      k            =            0                                n                                                              (                                            n            k                                              )                                                            (            −            1                          )              k                                            n            +            k            +            1                          .            where in step   (  ⋆  ) we use the substitution   k  →  n  −  k

Kristen Stokes · Accepted Answer

To evaluate the sum      ∑          k      =      0        n                                (                            n        k                              )                                (      −      1              )                  n          −          k                            n      +      k      +      1      introduce the function  f  (  z  )  =            n      !              n      +      1      +      z            ∏          q      =      0        n        1          z      −      q      which has the property that for   0  ≤  k  ≤  n            R      e      s              z      =      k        f  (  z  )  =            n      !              n      +      1      +      k            ∏          q      =      0              k      −      1            1          k      −      q            ∏          q      =      k      +      1        n        1          k      −      q          =            n      !              n      +      1      +      k            1          k      !                  (      −      1              )                  n          −          k                            (      n      −      k      )      !        =                              (                            n        k                              )                                (      −      1              )                  n          −          k                            n      +      1      +      k        .Now with residues summing to zero and the residue at infinity being zero by inspection we get for our sum the value  −            R      e      s              z      =      −      n      −      1        f  (  z  )  =  −  n  !      ∏          q      =      0        n        1          −      n      −      1      −      q          =  n  !  (  −  1      )          n            ∏          q      =      0        n        1          n      +      1      +      q        =  (  −  1      )          n        n  !            n      !              (      2      n      +      1      )      !        .This is                              (          −          1                      )            n                                    2          n          +          1                                                                                (                                                                    2                n                            n                                                      )                                                              −          1                    .

Denote the sum S n </msub> := <munderover> ∑ <mrow class

Answered question

Answer & Explanation

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