How to write sum of fractions with powersDisclaimer: I'm not a mathematician so this question may sound very easy to you. Sorry if this is too easy for you. (Also english is not my native language but I'll try.)I have the folowing formula: x ∗ 3 k ( 2 n ) k + 3 k − 1 ( 2 n ) k + 3 k − 2 ( 2 n ) k − 1 + . . . + 3 0 ( 2 n ) 1 k ∈ N , n ∈ N , x ∈ N The question seems simple to me: How do I / Is there another way to write this formula, so I don't have the dots in the middle?I you have any questions regarding this formula, please let me know, so I can clarify what I meant.Any help is appreciated

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How to write sum of fractions with powersDisclaimer: I&#039;m not a mathematician so this question may sound very easy to you. Sorry if this is too easy for you. (Also english is not my native language but I&#039;ll try.)I have the folowing formula:            x      ∗              3        k                    (      2      n              )        k              +            3              k        −        1                    (      2      n              )        k              +            3              k        −        2                    (      2      n              )                  k          −          1                      +  .  .  .  +            3              0                    (      2      n              )                  1                      k  ∈      N    ,    n  ∈      N    ,    x  ∈      N  The question seems simple to me: How do I / Is there another way to write this formula, so I don&#039;t have the dots in the middle?I you have any questions regarding this formula, please let me know, so I can clarify what I meant.Any help is appreciated

beentjie8e · Accepted Answer

You can write it as:            x      ∗              3        k                    (      2      n              )        k              +            3              k        −        1                    (      2      n              )        k              +            3              k        −        2                    (      2      n              )                  k          −          1                      +  .  .  .  +            3              0                    (      2      n              )                  1                      =            x      ∗              3        k                    (      2      n              )        k              +  (            3              k        −        1                    (      2      n              )        k              +            3              k        −        2                    (      2      n              )                  k          −          1                      +  .  .  .  +            3              0                    (      2      n              )                  1                      )  =            x      ∗              3        k                    (      2      n              )        k              +  (            3              k        −        1                    (      2      n              )        k              +            3              k        −        2                    (      2      n              )                  k          −          1                      +  .  .  .  +            3              i        −        1                    (      2      n              )                  i                      +  .  .  .  .  +            3              0                    (      2      n              )                  1                      )  =            x      ∗              3        k                    (      2      n              )        k              +      ∑          i      =      1        k              3              i        −        1                    (      2      n              )        i            In your original phrase &quot;            x      ∗              3        k                    (      2      n              )        k              +            3              k        −        1                    (      2      n              )        k              +            3              k        −        2                    (      2      n              )                  k          −          1                      +  .  .  .  +            3              0                    (      2      n              )                  1                    &quot; the first term does not match the rest of the others in patterns so it was isolated and written separately.====For the heck of it, we could write the             3              i        −        1                    (      2      n              )        i             terms as       1    3    (      3          2      n            )    i   so the whole thing is             x      ∗              3        k                    (      2      n              )        k              +      1    3        ∑          i      =      1        k    (      3          2      n            )    i   if that makes anything clearer, but we don&#039;t have to (and maybe it doesn&#039;t).      ∑          i      =      1        k    (      3          2      n            )    i   is a geometric series and you may (or may not know)       ∑          i      =      1        k    (      3          2      n            )    i    =            1      −      (              3                  2          n                            )                  k          +          1                            1      −              3                  2          n                     so the whole thing is             x      ∗              3        k                    (      2      n              )        k              +      1    3              1      −      (              3                  2          n                            )                  k          +          1                            1      −              3                  2          n                    But your question was about notation, not solving.In notatation you can write       a    k    +      a          k      −      1        +  .  .  .  .  .  +      a    1   as       ∑          i      =      1        k        a    i

wendi1019gt · Accepted Answer

As written, we can say            x      ⋅              3        k                    (      2      n              )        k              +            3              k        −        1                    (      2      n              )        k              +            3              k        −        2                    (      2      n              )                  k          −          1                      +  .  .  .  +            3              0                    (      2      n              )                  1                      =      (                  x        ⋅                  3          k                            (        2        n                  )          k                      )    +  Uwhere                    U                            =                              3                          k              −              1                                            (            2            n                          )              k                                      +                              3                          k              −              2                                            (            2            n                          )                              k                −                1                                                    +        .        .        .        +                              3                          0                                            (            2            n                          )                              1                                                                                      =                  1          3                                      3                          k                                            (            2            n                          )              k                                      +                  1          3                                      3                          k              −              1                                            (            2            n                          )                              k                −                1                                                    +        .        .        .        +                  1          3                                      3                          1                                            (            2            n                          )                              1                                                                                      =                  1          3                          (                                    3                              k                                                    (              2              n                              )                k                                              +                                    3                              k                −                1                                                    (              2              n                              )                                  k                  −                  1                                                              +          .          .          .          +                                    3                              1                                                    (              2              n                              )                                  1                                                              )                                                  =                  1          3                          (                                    (                              3                                  (                  2                  n                  )                                            )                        k                    +          …          +                                    (                              3                                  (                  2                  n                  )                                            )                        1                    )                    That middle thing is a geometric series, whose sum is                     3                  2          n                            k        +        1                    1      −              3                  2          n                      −  1, so that the whole thing is            x      ⋅              3        k                    (      2      n              )        k              )  +      1    3        (                            (                      3                          2              n                                )                          k          +          1                            1        −                  3                      2            n                                −    1    )    .

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