Prove n − 1 n + n − 1 n n − 3 n − 2 + n − 1 n n − 3 n − 2 n − 5 n − 4 + . . . = n − 1 3 without inductionI found this identity using Maple. Is there a (simple) way to prove it without using induction? Using induction, the proof is quite easy.Prove for odd n that ∑ k = 1 ( n + 1 ) / 2 ∏ j = 0 k − 1 ( n − 2 j − 1 n − 2 j ) = n − 1 n + n − 1 n n − 3 n − 2 + n − 1 n n − 3 n − 2 n − 5 n − 4 + . . . = n − 1 3

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Prove             n      −      1        n    +            n      −      1        n              n      −      3              n      −      2        +            n      −      1        n              n      −      3              n      −      2                  n      −      5              n      −      4        +  .  .  .  =            n      −      1        3   without inductionI found this identity using Maple. Is there a (simple) way to prove it without using induction? Using induction, the proof is quite easy.Prove for odd n that      ∑          k      =      1              (      n      +      1      )              /            2            ∏          j      =      0              k      −      1            (                  n        −        2        j        −        1                    n        −        2        j              )    =            n      −      1        n    +            n      −      1        n              n      −      3              n      −      2        +            n      −      1        n              n      −      3              n      −      2                  n      −      5              n      −      4        +  .  .  .  =            n      −      1        3

Skye Hamilton · Accepted Answer

We need to prove that            n      −      1        n    +            n      −      1        n              n      −      3              n      −      2        +            n      −      1        n              n      −      3              n      −      2                  n      −      5              n      −      4        +  .  .  .  =            n      −      1        3  or  1  +            n      −      3              n      −      2        +            n      −      3              n      −      2                  n      −      5              n      −      4        +  .  .  .  =      n    3  or            n      −      3              n      −      2        +            n      −      3              n      −      2                  n      −      5              n      −      4        +  .  .  .  =            n      −      3        3  or  1  +            n      −      5              n      −      4        +            n      −      5              n      −      4                  n      −      7              n      −      6        .  .  .  =            n      −      2        3  or  1  +      2    3    =      5    3    .Done!

Prove (n-1)/(n)+(n-1)/(n)(n-3)/(n-2)+ (n-1)/(n)(n-3)/(n-2)(n-5)/(n-4) + ... = (n-1)/(3) without induction

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