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**Soroban** Hello, Grillakis!

Your work is excellent!

We know that $\displaystyle f(k)$ is divisible by 6: .$\displaystyle f(k) = 6a,\:\text{ for some integer }a.$

$\displaystyle \text{So we have: }\;f(k+1)\:=\:6a + 3\underbrace{k(k+1)}$

Note that $\displaystyle k(k+1)$ is the product of *two consecutive integers.*

Hence, one of them is even and the other is odd.

And their product is even, say, $\displaystyle 2b,\:\text{ for some integer }b.$

So we have: .$\displaystyle f(k+1)\:=\:6a + 3(2b) \:=\:6(a + b)\;\hdots\;\text{ which is divisible by 6.}$