\((15, 33, 57, 87, 123, \ldots)\text{.}\) Try dividing each term by 3. That gives the sequence \(5, 11, 19, 29, 41,\ldots\text{.}\) Now add 1 to each term: \(6, 12, 20, 30, 42, \ldots\text{,}\) which is \((d_n)\) in this example, except starting with 6 instead of 0. So let's start with the formula \(d_n= n(n+1)\text{.}\) To start with the 6, we shift: \((n+2)(n+3)\text{.}\) But this is one too many, so subtract 1: \((n+2)(n+3) - 1\text{.}\) That gives us our sequence, but divided by 3. So we want \(j_n = 3((n+2)(n+3) - 1)\text{.}\)
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