How to calculate ∫ Q ( x 1 + x 2 + … + x n ) 2 d λ n whereas n ≥ 2 and Q = { ( x 1 , … , x n ) ∈ R n : 0 ≤ x i ≤ 1 , i = 1 , … , n } I know that I have to use induction and Fubini to calculate I ( n ) = ∫ 0 1 ∫ 0 1 ⋯ ∫ 0 1 ( x 1 + x 2 + … + x n ) 2 d x n ⋯ d x 2 d x 1 And I already started to calculate the integrals for n = 2 , 3 , 4 but the numbers I got did not enlighten me and I have no idea for an induction hypothesis. I also exchanged x with 1 but I still have no clue. Maybe I am doing something wrong. Thanks for help in advance.

Question

How to calculate      ∫          Q                  (              x                  1                    +              x                  2                    +      …      +              x                  n                    )              2        d      λ          n      whereas   n  ≥  2 and  Q  =      {          (              x                  1                    ,      …      ,              x                  n                    )        ∈                  R                    n              :    0    ≤          x              i              ≤    1    ,    i    =    1    ,    …    ,    n    }  I know that I have to use induction and Fubini to calculate  I  (  n  )  =      ∫    0    1        ∫    0    1    ⋯      ∫    0    1              (              x                  1                    +              x                  2                    +      …      +              x                  n                    )              2        d      x    n    ⋯    d      x    2      d      x    1  And I already started to calculate the integrals for   n  =  2  ,  3  ,  4 but the numbers I got did not enlighten me and I have no idea for an induction hypothesis. I also exchanged   x with 1 but I still have no clue. Maybe I am doing something wrong. Thanks for help in advance.

Blaze Frank · Accepted Answer

By expanding  (      x    1    +  ⋯  +      x    n        )    2    =      ∑          i      =      1        n        x    i    2    +  2      ∑          1      ≤      i      &amp;lt;      j      ≤      n            x    i        x    j  we see that  I  (  n  )  =      ∑          i      =      1        n        1    3    +  2      ∑          1      ≤      i      &amp;lt;      j      ≤      n            1    2    ⋅      1    2    =      n    3    +      1    2    ⋅            n      (      n      −      1      )        2    =            n      (      3      n      +      1      )        12  Alternatively, use the identity  (      x    1    +  ⋯  +      x    n        )    2    =  (      x    1    +  ⋯  +      x          n      −      1            )    2    +  2      x    n    (      x    1    +  ⋯  +      x          n      −      1        )  +      x    n    2  to obtain the recursion  I  (  n  )  =  I  (  n  −  1  )  +            n      −      1        2    +      1    3  Since   I  (  1  )  =  1      /    3, then  I  (  n  )  =      1    3    +      ∑          j      =      1              n      −      1            (          j      2        +          1      3        )  which evaluates to   n  (  3  n  +  1  )      /    12.

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