A few days ago
math help pre calc HELP ME!!!!!!!1?
prove that for any positive integer “n” we have that
n^3 + 11n is divisible by 6
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A few days ago
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Use an inductive method: first show it works for 1:1^3 + 11(1) = 12 which works
Then assume it works for n, and show it works for (n+1)
(n+1)^3 + 11(n+1) = n^3 + 3n^2 + 3n + 1 + 11n + 11
= n^3 + 3n^2 + 14n + 12
Since you assumed n^3 + 11n is divisible by 6 you can subtract it and see if what is left is divisible by 6
It would be 3n^2 + 3n + 12
This is 3n(n+1) + 12
Subtract the 12 since it’s a multiple of 6
3n(n+1) Multiples of 6 are any numbers that are both multiples of 2 AND 3. Clearly this is a multiple of 3, and it must also be a multiple of 2 since n(n+1), the product of two consecutive integers, is always even
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