Proving a binomial sum identity ∑ k = 0 n ( n k ) ( − 1 ) k 2 k + 1 = ( 2 n ) ! ! ( 2 n + 1 ) ! ! .

Question

Proving a binomial sum identity       ∑          k      =      0        n                  (              n      k                  )                  (      −      1              )        k                    2      k      +      1        =            (      2      n      )      !      !              (      2      n      +      1      )      !      !        .

Sanaa Hinton · Accepted Answer

You could consider the integral      ∫      0      1    (  1  −      x    2        )    n    d  x  .

EnvivyEvoxys6 · Accepted Answer

S    n    =      ∑          k      =      0        n                      (        −        1                  )          k                                                (                                n            k                                          )                                      2        k        +        1            We have:                              S          n                                    =                                            (              −              1                              )                n                                                    2              n              +              1                                      +                  ∑                      k            =            0                                n            −            1                                                              (              −              1                              )                k                                                                              (                                                  n                  k                                                                  )                                                                    2              k              +              1                                                                        =                                            (              −              1                              )                n                                                    2              n              +              1                                      +                  ∑                      k            =            0                                n            −            1                                    [                                    n                              n                −                k                                              .                                                    (                −                1                                  )                  k                                                                                        (                                                                              n                      −                      1                                        k                                                                          )                                                                              (                2                k                +                1                )                                              ]                                                  =                                            (              −              1                              )                n                                                    2              n              +              1                                      +                              1                          2              n              +              1                                                ∑                      k            =            0                                n            −            1                                    [                                                    n                (                2                n                +                1                )                                            n                −                k                                              .                                                    (                −                1                                  )                  k                                                                                        (                                                                              n                      −                      1                                        k                                                                          )                                                                              (                2                k                +                1                )                                              ]                                                  =                                            (              −              1                              )                n                                                    2              n              +              1                                      +                              1                          2              n              +              1                                                ∑                      k            =            0                                n            −            1                                    [                                                    2                                  n                  2                                −                2                n                k                +                2                n                k                +                n                                            n                −                k                                              .                                                    (                −                1                                  )                  k                                                                                        (                                                                              n                      −                      1                                        k                                                                          )                                                                              (                2                k                +                1                )                                              ]                                                  =                                            (              −              1                              )                n                                                    2              n              +              1                                      +                              1                          2              n              +              1                                                ∑                      k            =            0                                n            −            1                                    [                                                    2                                  n                  2                                −                2                n                k                                            n                −                k                                              .                                                    (                −                1                                  )                  k                                                                                        (                                                                              n                      −                      1                                        k                                                                          )                                                                              (                2                k                +                1                )                                              ]                                                  +                              1                          2              n              +              1                                                ∑                      k            =            0                                n            −            1                                    [                                                    2                n                k                +                n                                            n                −                k                                              .                                                    (                −                1                                  )                  k                                                                                        (                                                                              n                      −                      1                                        k                                                                          )                                                                              (                2                k                +                1                )                                              ]                                                  =                                            (              −              1                              )                n                                                    2              n              +              1                                      +                                            2              n                                      2              n              +              1                                                ∑                      k            =            0                                n            −            1                                                              (              −              1                              )                k                                                                              (                                                                      n                    −                    1                                    k                                                                  )                                                                    (              2              k              +              1              )                                      +                              1                          2              n              +              1                                                ∑                      k            =            0                                n            −            1                                                              n              (              −              1                              )                k                                                                              (                                                                      n                    −                    1                                    k                                                                  )                                                                    n              −              k                                                                        =                                            2              n                                      2              n              +              1                                                ∑                      k            =            0                                n            −            1                                                              (              −              1                              )                k                                                                              (                                                                      n                    −                    1                                    k                                                                  )                                                                    (              2              k              +              1              )                                      +                              1                          2              n              +              1                                                ∑                      k            =            0                                n            −            1                                    [          (          −          1                      )            k                                                          (                                      n              k                                                  )                                ]                +                                            (              −              1                              )                n                                                    2              n              +              1                                                                        =                                            2              n                                      2              n              +              1                                                ∑                      k            =            0                                n            −            1                                                              (              −              1                              )                k                                                                              (                                                                      n                    −                    1                                    k                                                                  )                                                                    (              2              k              +              1              )                                      +                              1                          2              n              +              1                                                ∑                      k            =            0                    n                          [          (          −          1                      )            k                                                          (                                      n              k                                                  )                                ]                    Therefore,      S    n    =                    2        n                    2        n        +        1                  S          n      −      1        +  0  ⇒      S          n      −      1        =                    2        n        −        2                    2        n        −        1                  S          n      −      2        .  .  .  ⇒      S    1    =            2      3            S    0   and       S    0    =  1Hence,      S    n    =                    2        n        (        2        n        −        2        )        .        .        .2                    (        2        n        +        1        )        (        2        n        −        1        )        .        .        .3        .1              =                    (        2        n        )        !        !                    (        2        n        +        1        )        !        !

Proving a binomial sum identity <munderover> ∑ k =

Answered question

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