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Consecutive Cube
Problem
Three consecutive integers are multiplied together and the middle number is added.
E.g. 3 
4 5 = 60 and 60 + 4 = 64 = 43
Will this always produce a cube number?
Solution
Taking three consecutive integers, n, n + 1 and n + 2 we get.
n(n + 1)(n + 2) + (n + 1) | = (n + 1)(n(n + 2) + 1) |
| = (n + 1)(n2 + 2n + 1) | |
| = (n + 1)(n + 1)2 | |
| = (n + 1)3 |
Problem ID: 83 (Oct 2002) Difficulty: 2 Star
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