Since everyone freaked out, I made the variables are the same. ∑ x = 1 n 2 x − 1

Question

Since everyone freaked out, I made the variables are the same.      ∑          x      =      1              n            2          x      −      1

Ryan Newman · Accepted Answer

In this instance, without explicitly using the formula for geometric series,                              ∑                      x            =            1                    n                          2                      x            −            1                                              =        1        +        2        +                  2          2                +                  2          3                +        ⋯        +                  2                      n            −            1                                                            =        1        +        (        1        +        2        +                  2          2                +                  2          3                +        ⋯        +                  2                      n            −            1                          )        −        1                    add and subtract                 1                                          =        (        1        +        1        )        +        (        2        +                  2          2                +                  2          3                +        ⋯        +                  2                      n            −            1                          )        −        1                    regroup                                          =        2        +        (        2        +                  2          2                +                  2          3                +        ⋯        +                  2                      n            −            1                          )        −        1                                          =        (        2        +        2        )        +        (                  2          2                +        ⋯        +                  2                      n            −            1                          )        −        1                    regroup again                                          =                  2          2                +        (                  2          2                +                  2          3                +        ⋯        +                  2                      n            −            1                          )        −        1                                          =        (                  2          2                +                  2          2                )        +        (                  2          3                +        ⋯        +                  2                      n            −            1                          )        −        1                    regroup again                                          =                  2          3                +        (                  2          3                +        ⋯        +                  2                      n            −            1                          )        −        1                                          =        ⋯                    lather, rinse, repeat                                          =                  2                      n            −            1                          +        (                  2                      n            −            1                          )        −        1                    nearly done                                          =                  2          n                −        1.            Now that we know the form of the result, it is also possible to prove the result      ∑          x      =      1        n        2          x      −      1        =      2    n    −  1more formally by induction. Clearly, the result holds when n=1 since       2    0    =      2    1    −  1. Then, if the result holds for some positive integer n, we have that      ∑          x      =      1              n      +      1            2          x      −      1        =      ∑          x      =      1        n        2          x      −      1        +      2    n    =  (      2    n    −  1  )  +      2    n    =      2          n      +      1        −  1and so the result holds for n+1 as well. Since we know that the result holds when n=1, it follows by induction that it holds for all positive integers n.

Cory Patrick · Accepted Answer

You&#039;re saying you want as outputs  1  ,  3  ,  7  ,  15  ,  31  ,  63Note they are respectively       2    1    −  1  ,      2    2    −  1  ,      2    3    −  1  ,      2    4    −  1  ,      2    5    −  1  ,      2    6    −  1 so what you really want is  f  (  n  )  =      2    n    −  1Now this is a finite geometric sum, namely      ∑          i      =      0              n      −      1            2    i    =      2    n    −  1Now this is a finite geometric sum, namely      ∑          i      =      0              n      −      1            2    i    =      2    n    −  1      ∑          i      =      0              n      −      1            2    i    =      2    n    −  1This follows from the geometric sum formula, that is      ∑          i      =      0              n      −      1            a    i    =                    a        n            −      1              a      −      1      The MO for this is the following. Let our sum be S  1  +  a  +  ⋯  +            a              n        −        1              =  SThen  a  +            a      2        +  ⋯  +            a      n        =  a  SBut  a  +            a      2        +  ⋯  +            a      n        =      (          1      +      a      +      ⋯      +                        a                      n            −            1                                )    −  1  +            a      n        =  S  −  1  +            a      n      So that                          S        −        1        +                              a            n                          =        a        S                                  S        −        a        S        =        1        −                              a            n                                                              (                      1            −            a                    )                S        =        1        −                              a            n                                                    S        =                              1            −                                          a                n                                                          1            −            a                              as desired.

Since everyone freaked out, I made the variables are the same. <munderover> ∑<!-- ∑ -

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