1 2 ⋅ 4 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + 1 ⋅ 3 ⋅ 5 ⋅ 7 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 + ⋯is equal to?

Question

1          2      ⋅      4        +            1      ⋅      3              2      ⋅      4      ⋅      6        +            1      ⋅      3      ⋅      5              2      ⋅      4      ⋅      6      ⋅      8        +            1      ⋅      3      ⋅      5      ⋅      7              2      ⋅      4      ⋅      6      ⋅      8      ⋅      10        +  ⋯is equal to?

wasipewelr · Accepted Answer

If you look at the Binomial expansion of  (  1  −  x      )          −              1        2            you get :-      ∑          r      =      0              ∞                                                (                                                2            r                    r                                      )                                      x                  r                            4              r            So      ∫          0              1        (  1  −  x      )          −              1        2              d  x  =      ∑          r      =      0              ∞                                    (                                      2          r                r                              )                                    4                  r                    (      r      +      1      )      So you get      1    2        ∑          r      =      0              ∞                                    (                                      2          r                r                              )                                    4                  r                    (      r      +      1      )        =      1    2        ∑          r      =      0              ∞                  (      2      r      )      !                      4                  r                    r      !      (      r      +      1      )      !        =      1    2        ∫          0              1        (  1  −  x      )          −              1        2              d  x  =  1So      1    2        ∑          r      =      1              ∞                  (      2      r      )      !                      4                  r                    r      !      (      r      +      1      )      !        =      1    2        ∑          r      =      0              ∞                  (      2      r      )      !                      4                  r                    r      !      (      r      +      1      )      !        −      1    2    (  1  )  =  1  −      1    2    =      1    2

Wisniewool · Accepted Answer

I tried to rewrite the sum:      ∑          n      =      1        k                                (                      1            4                    )                n            (      2      n      )      !              n      !      (      n      +      1      )      !      As a result I got:                    2                  −          2          k          −          1                            (        −        k        (        2        (        k        +        1        )        )        !        −        2        (        2        (        k        +        1        )        )        !        +                  2                      2            k            +            1                          (        k        +        1        )        !        (        k        +        2        )        !        )                    (      k      +      1      )      !      (      k      +      2      )      !          =  1  −                    2                  −          2          k          −          1                    (      k      +      2      )      (      2      (      k      +      1      )      )      !              (      k      +      1      )      !      (      k      +      2      )      !        =  1  −            2      Γ              (        k        +                  3          2                )                            π            Γ      (      k      +      2      )      where   Γ is Euler Gamma Function. By setting   k  =  ∞, the complete sum becomes 1.

1 2 ⋅ 4 </mrow> </mfrac> + 1

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