The sum of any three consecutive odd numbers is proved to be 11 more than a multiple of 12.
In the question, we are asked to prove that the sum of the squares of any three consecutive odd numbers is always 11 more than a multiple of 12.
We assume n to be any whole number.
Thus, we choose our 3 consecutive odd numbers as:
2n + 1, 2n + 3, and 2n + 5.
The sum of the squares of these numbers can be shown as:
(2n + 1)² + (2n + 3)² + (2n + 5)²
= (4n² + 4n + 1) + (4n² + 12n + 9) + (4n² + 20n + 25) {Using (a + b)² = a² + 2ab + b²}
= 12n² + 36n + 35
= 12n² + 36n + 24 + 11
= 12(n² + 3n + 2) + 11
= 12k + 11, where n² + 3n 2 is taken as k, any whole number as n is a whole number.
Thus, the sum is 11 more than a multiple of 12, that is, 12k.
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