\((6, 10, 18, 34, 66, \ldots )\text{.}\) To get from one term to the next, we almost double each term. So maybe we can relate this back to \(2^n\text{.}\) Yes, each term is 2 more than a power of 2. So we get \(h_n = 2^{n+2} + 2\) (the \(n+2\) is because the first term is 2 more than \(2^2\text{,}\) not \(2^0\)). Alternatively, we could have related this sequence to the second sequence in this example: starting with 3, 5, 9, 17, … we see that this sequence is twice the terms from that sequence. That sequence had closed formula \(c_n = 2^{n+1} + 1\text{.}\) Our sequence here would be twice this, so \(h_n = 2(2^n + 1)\text{,}\) which is the same as we got before.
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