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Permutations And Combinations
Let n + 1, n + 2, .....(n + m) be m consecutive integers
Let denote the sum of the cubes of these integers and denote the sum of these integers
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= [1 + 2 + 3 +........ + n + (n + 1) + ..... + (n + m)]  [1 + 2 + ....... + n]
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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Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.
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