Permutations And Combinations

Question
CBSEENMA11013177

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

Solution

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.
   

Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.