How could one prove that ∑ k = 0 ∞ 2 1 − k ( 3 − 25 k ) ( 2 k ) ! k ! ( 3 k ) ! = − π

Question

How could one prove that      ∑          k      =      0        ∞                      2                  1          −          k                    (      3      −      25      k      )      (      2      k      )      !            k      !              (      3      k      )      !        =  −  π

tezeuszxcL · Accepted Answer

Use Beta function, I guess... for   k  ≥  1      ∫    0    1        t          2      k        (  1  −  t      )          k      −      1        d  t  =  B  (  k  ,  2  k  +  1  )  =            (      k      −      1      )      !      (      2      k      )      !              (      3      k      )      !      So write  f  (  x  )  =      ∑          k      =      0        ∞              2      (      3      −      25      k      )      k      !      (      2      k      )      !              (      3      k      )      !            x    k  and compute   f  (  1      /    2  ) like this:                    f        (        x        )                            =        6        +                  ∑                      k            =            1                    ∞                                      (            6            −            50            k            )            k            (            k            −            1            )            !            (            2            k            )            !                                (            3            k            )            !                                    x          k                                                  =        6        +                  ∑                      k            =            1                    ∞                (        6        −        50        k        )        k                  x          k                          ∫          0          1                          t                      2            k                          (        1        −        t                  )                      k            −            1                          d        t                                          =        6        +                  ∫          0          1                          ∑                      k            =            1                    ∞                (        6        −        50        k        )        k                  x          k                          t                      2            k                          (        1        −        t                  )                      k            −            1                                  d        t                                          =        6        +                  ∫          0          1                                      4                          t              2                        x            (            14                          t              3                        x            −            14                          t              2                        x            −            11            )                                (                          t              3                        x            −                          t              2                        x            +            1                          )              3                                              d        t                            f                  (                      1            2                    )                                    =        6        +                  ∫          0          1                                      16                          t              2                        (            7                          t              3                        −            7                          t              2                        −            11            )                                (                          t              3                        −                          t              2                        +            2                          )              3                                              d        t        =        −        π

How could one prove that sum_(k=0)^infty (2^(1-k)(3-25k)(2k)!k!)/(3k)!=-pi

Answered question

Answer & Explanation

New Questions in Force, Motion and Energy