Find 1 ⋅ 2 3 ! + 2 ⋅ 2 2 4 ! + 3 ⋅ 2 3 5 ! + 4 ⋅ 2 4 6 ! + ⋯ up to n terms?

Question

Find                               1          ⋅          2                          3          !                    +                        2          ⋅                      2            2                                    4          !                    +                        3          ⋅                      2            3                                    5          !                    +                        4          ⋅                      2            4                                    6          !                    +      ⋯       up to n terms?

aletantas1x · Accepted Answer

So:                              ∑                      k            =            1                    n                                                    k                              2                k                                                    (              k              +              2              )              !                                                          =                  ∑                      k            =            1                    n                                                    (              k              +              2              )                              2                k                            −                              2                                  k                  +                  1                                                                    (              k              +              2              )              !                                      =                  ∑                      k            =            1                    n                                                                    2                k                                                    (              k              +              1              )              !                                      −                  ∑                      k            =            1                    n                                                                    2                                  k                  +                  1                                                                    (              k              +              2              )              !                                                                        =                  ∑                      k            =            1                    n                                                                    2                k                                                    (              k              +              1              )              !                                      −                  ∑                      k            =            2                                n            +            1                                                                              2                k                                                    (              k              +              1              )              !                                      =        1        −                                            2                              n                +                1                                                          (            n            +            2            )            !                          .

varitero5w · Accepted Answer

Making the problem more general, write      ∑          m      =      1        n              m                    x        m                    (      m      +      2      )      !        =      1    x        ∑          m      =      1        n              x              m        +        1                    (      m      +      1      )      !        −      2          x      2            ∑          m      =      1        n              x              m        +        2                    (      m      +      2      )      !            ∑          m      =      1        n              x              m        +        1                    (      m      +      1      )      !        =      e    x              Γ      (      n      +      2      ,      x      )              (      n      +      1      )      !        −  x  −  1      ∑          m      =      1        n              x              m        +        2                    (      m      +      2      )      !        =      e    x              Γ      (      n      +      3      ,      x      )              Γ      (      n      +      3      )        −            x      2        2    −  x  −  1Recombining everything      ∑          m      =      1        n              m                    x        m                    (      m      +      2      )      !        =            x      +      2              x      2        −            x              n        +        1                    (      n      +      2      )      !        +      e    x              x      −      2              x      2                  Γ      (      n      +      3      ,      x      )              (      n      +      2      )      !      You are lucky with   x  =  2      ∑          m      =      1        n              m                    2        m                    (      m      +      2      )      !        =  1  −            2              n        +        1                    (      n      +      2      )      !

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