Permutations and Combinations

Permutations and Combinations

Question

Show that the sum of the cubes of any number of consecutive integers is divisible by the sum of these integers.

Answer

Let n + 1, n + 2, .....(n + m) be m consecutive integers
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#6 {main}</pre>  denote the sum of the cubes of these integers andspace space straight S subscript 1 denote the sum of these integers
space space straight S subscript 3 space equals space left parenthesis straight n plus 1 right parenthesis cubed plus left parenthesis straight n plus 2 right parenthesis cubed plus......... plus left parenthesis straight n plus straight m right parenthesis cubed

      =  left square bracket 1 cubed plus 2 cubed plus.......... plus straight n cubed plus left parenthesis straight n plus 1 right parenthesis cubed plus.......... plus left parenthesis straight n plus straight m right parenthesis cubed right square bracket space minus space left square bracket 1 cubed plus 2 cubed plus..... straight n cubed right square bracket
      = sum from straight k equals 1 to straight n plus straight m of straight k cubed space minus space sum from straight k equals 1 to straight n of straight k cubed                                                  space space space space space space space space open square brackets because space space sum from straight k equals 1 to straight n of straight k cubed space equals space fraction numerator straight n squared left parenthesis straight n plus 1 right parenthesis squared over denominator 4 end fraction close square brackets
      = fraction numerator left parenthesis straight n plus straight m right parenthesis squared left parenthesis straight n plus straight m plus 1 right parenthesis squared over denominator 4 end fraction minus fraction numerator straight n squared left parenthesis straight n plus 1 right parenthesis squared over denominator 4 end fraction
      = open square brackets fraction numerator left parenthesis straight n plus straight m right parenthesis left parenthesis straight n plus straight m plus 1 right parenthesis over denominator 2 end fraction minus fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets space cross times space open square brackets fraction numerator left parenthesis straight n plus straight m right parenthesis space left parenthesis straight n plus straight m plus 1 right parenthesis over denominator 2 end fraction plus fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets
straight S subscript 1 space equals space left parenthesis straight n space plus space 1 right parenthesis space plus space left parenthesis straight n space plus space 2 right parenthesis space plus space.......... space plus space left parenthesis straight n space plus space straight m right parenthesis
    = [1 + 2 + 3 +........ + n + (n + 1) + ..... + (n + m)] - [1 + 2 + ....... + n]
   space space space space equals space sum from straight k equals 1 to straight n plus straight m of straight k space minus space sum from straight k equals 1 to straight n of straight k space equals space fraction numerator left parenthesis straight n plus straight m right parenthesis space left parenthesis straight n plus straight m plus 1 right parenthesis over denominator 2 end fraction space minus space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction          open square brackets because space space sum from straight k equals 1 to straight n of straight k space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets
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#6 {main}</pre>

         equals space space fraction numerator left parenthesis straight n plus straight m right parenthesis left parenthesis straight n plus straight m plus 1 right parenthesis over denominator 2 end fraction space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction
which is an integer because (n + m) (n + m + 1) and n(n + 1) are both even numbers being the product of two consecutive integers. Hence, the sum of cubes of any number of consecutive integers is divisible by the sum of these integers.
   

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