[ n 1 ] + [ n 2 ] + ⋯ + [ n n ] + [ n ] is evenI tried this by taking two different cases, when, n is even and when n is odd. But I could not establish any relation between the two. Please help to solve this problem and thanks a lot in advance.P.S. Although nothing is mentioned in the question I think [ ] stands for greatest integer function as otherwise the sum would be in a fraction which is not even as only natural numbers are regarded as even.

Question

[          n      1        ]    +      [          n      2        ]    +  ⋯  +      [          n      n        ]    +      [          n        ]   is evenI tried this by taking two different cases, when,   n is even and when   n is odd. But I could not establish any relation between the two. Please help to solve this problem and thanks a lot in advance.P.S. Although nothing is mentioned in the question I think   [     ] stands for greatest integer function as otherwise the sum would be in a fraction which is not even as only natural numbers are regarded as even.

Bruno Hughes · Accepted Answer

Use the fact that:      ∑          i      =      1        n        ⌊          n      i        ⌋    =      ∑          i      =      1        n    d  (  i  )where   d  (     ) is the divisor function (RHS: there are   d  (  i  ) numbers dividing   i, LHS: there are       ⌊          n      i        ⌋   numbers at most   n divisible by   i). All numbers have even number of divisors, except for the squares, and there are   ⌊      n    ⌋ of those, thus:  ⌊      n    ⌋  +      ∑          i      =      1        n        ⌊          n      i        ⌋    =      ∑          i      =      1        n    d  (  i  )  +  1  {  ∃  k  ∈      N    :  i  =      k    2    }And all terms in the sum on the RHS are even.

[ n 1 </mfrac> ] </mrow> + [ n 2 </

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