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I am to derive the formula for $15,66,189,420, 795,\dots$

I have the common pattern as $3m(2n+1)$ where $n$ and $m$ are integers. For example,

$66=3\cdot2\cdot(2\cdot5+1)$

Will this end up in me deriving the correct formula or is there another pattern? Also could someone help me with the next step.

amWhy
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Kadoe
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1 Answers1

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You seem to be on the right track. The numbers are

$$3\cdot 1\cdot 5$$

$$3\cdot 2\cdot 11$$

$$3\cdot 3\cdot 21$$

$$3\cdot 4\cdot 35$$

$$3\cdot 5\cdot 53$$

The differences in the last column are $6,10,14,18$ , forming an arithmetic progression. So, we have $$a_n=3n\cdot (2n^2+3)=6n^3+9n$$

Peter
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