If f ( n ) = [ n + 1 2 ] when n is a natural number and [⋅] represents the floor function, then find the value of ∑ n = 1 ∞ 2 f ( n ) + 2 − f ( n ) 2 n

Question

If   f  (  n  )  =      [          n        +                  1        2              ]   when n is a natural number and [⋅] represents the floor function, then find the value of      ∑          n      =      1              ∞                                    2                      f            (            n            )                          +                  2                      −            f            (            n            )                                      2        n

Ryan Fitzgerald · Accepted Answer

Note : if   k  ∈            N              ∗      , then       (    ∀    p    ∈          [                          [                  k                      2                          −        k        +        1        ,                  k                      2                          +        k        ]                        ]        )    ,     f      (    p    )    =  k                              ∑                      n            =            1                                +            ∞                                                                              2                                  f                                      (                    n                    )                                                              +                              2                                  −                  f                                      (                    n                    )                                                                                      2                              n                                                                        =                  ∑                      k            =            1                                +            ∞                                                (                          ∑                              n                =                1                                                                                  (                    k                    +                    1                    )                                                        2                                                  −                                  (                  k                  +                  1                  )                                                                                                                          2                                          f                                              (                        n                        )                                                                              +                                      2                                          −                      f                                              (                        n                        )                                                                                                              2                                      n                                                                        −                          ∑                              n                =                1                                                              k                                      2                                                  −                k                                                                                                          2                                          f                                              (                        n                        )                                                                              +                                      2                                          −                      f                                              (                        n                        )                                                                                                              2                                      n                                                                        )                                                            =                  ∑                      k            =            1                                +            ∞                                                ∑                          n              =                              k                                  2                                            −              k              +              1                                                      k                                  2                                            +              k                                                                                            2                                      f                                          (                      n                      )                                                                      +                                  2                                      −                    f                                          (                      n                      )                                                                                                  2                                  n                                                                                                      =                  ∑                      k            =            1                                +            ∞                                                (                          2                              k                                      +                          2                              −                k                                      )                                ∑                          n              =                              k                                  2                                            −              k              +              1                                                      k                                  2                                            +              k                                                          1                              2                                  n                                                                                                      =                  ∑                      k            =            1                                +            ∞                                                2                          −              k                              (                k                +                1                )                                                          (                          2                              k                                      +                          2                              −                k                                      )                                (                          2                              2                k                                      −            1            )                                                            =                  ∑                      k            =            1                                +            ∞                                                (                          2                              2                k                −                                  k                                      2                                                                        −                          2                              −                2                k                −                                  k                                      2                                                                        )                                                            =                  ∑                      k            =            1                                +            ∞                                                (                          2                              −                k                                  (                  k                  −                  2                  )                                                      −                          2                              −                                  (                  k                  +                  1                  )                                                  (                  k                  −                  1                  )                                                      )                          +                  ∑                      k            =            1                                +            ∞                                                (                          2                              −                                  (                  k                  +                  1                  )                                                  (                  k                  −                  1                  )                                                      −                          2                              −                                  (                  k                  +                  2                  )                                k                                      )                                                            =                  2                      −            1            ×                          (              1              −              2              )                                      +                  2                      −                          (              1              +              1              )                        ×                          (              1              −              1              )                                                                    ∑                      n            =            1                                +            ∞                                                                              2                                  f                                      (                    n                    )                                                              +                              2                                  −                  f                                      (                    n                    )                                                                                      2                              n                                                                        =        3

If f ( n ) = [ <msqrt> n </msqrt> + <mstyle displaystyle="tr

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