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**GrigOrig99** Having a bit of trouble with this one. The final part of my attempt, highlighted in black, is meant to be factored in such a way that it ties into the earlier part of **1**. I've tried factoring the final part in different ways, but I just can't get it to match **1**. Can anyone help?

Many thanks.

Q. n(n + 1)(2n + 1) is divisible by 3 for 3 for n $\displaystyle \in \mathbb{N}_0$.

Attempt: __Step 1:__ For n = 1...

1(1 + 1)(2(1) + 1) => 1(2)(3) => 6, which can be divided by 3.

Therefore, n = 1 is true.

__Step 2:__ For n = k...

Assume k(k + 1)(2k + 1) can be divided by 3,

i.e. k(k + 1)(2k + 1) = 3Z, where Z is an integer...**1**

Show that n = k + 1 is true...

i.e. (k + 1)(k + 2)(2(k + 1) + 1) can be divided by 3.

(k + 1)(k + 2)(2(k + 1) + 1) => (k + 1)(k + 2)(2k + 2 + 1) => (k + 1)(k + 2)(2k + 3) =>

**2k(k + 1)(k + 2) + 3(k + 1)(k + 2)**