We wish to compare these sequences to the triangular numbers \((0, 1, 3, 6, 10, 15, 21,\ldots)\text{,}\) when we start with \(n=0\text{,}\) and the powers of 2: \((1, 2, 4, 8, 16, \ldots)\text{.}\)
\((1, 2, 4, 7, 11, 16, 22, \ldots)\text{.}\) Note that if subtract 1 from each term, we get the sequence \((T_n)\text{.}\) So we have \(b_n = T_n + 1\text{.}\) Therefore a closed formula is \(b_n = \frac{n(n+1)}{2} + 1\text{.}\) A quick check of the first few \(n\) confirms we have it right.
\((3, 5, 9, 17, 33, \ldots )\text{.}\) Each term in this sequence is one more than a power of 2, so we might guess the closed formula is \(c_n = a_n+1 = 2^n + 1\text{.}\) If we try this though, we get \(c_0 = 2^0 + 1 = 2\) and \(c_1 = 2^1 + 1 = 3\text{.}\) We are off because the indices are shifted. What we really want is \(c_n = a_{n+1}+1\) giving \(c_n = 2^{n+1} + 1\text{.}\)
(\(0, 2, 6, 12, 20, 30, 42,\ldots \)). Notice that all these terms are even. What happens if we factor out a 2? We get \((T_n)\text{!}\) More precisely, we find that \(d_n/2 = T_n\text{,}\) so this sequence has closed formula \(d_n = n(n+1)\text{.}\)
\((3, 6, 10, 15, 21, 28, \ldots)\text{.}\) These are all triangular numbers. However, we are starting with 3 as our initial term instead of as our third term. So if we could plug in 2 instead of 0 into the formula for \(T_n\text{,}\) we would be set. Therefore the closed formula is \(e_n = \frac{(n+2)(n+3)}{2}\) (where \(n+3\) came from \((n+2)+1\)). Thinking about sequences as functions, we are doing a horizontal shift by 2: \(e_n = T_{n+2}\) which would cause the graph to shift 2 units to the left.
\((0, 1, 3, 7, 15, 31, \ldots )\text{.}\) Try adding 1 to each term and we get powers of 2. You might guess this because each term is a little more than twice the previous term (the powers of 2 are exactly twice the previous term). Closed formula: \(f_n = 2^{n} - 1\text{.}\)
\((3, 6, 12, 24, 48, \ldots )\text{.}\) These numbers are also doubling each time, but are also all multiples of 3. Dividing each by 3 gives 1, 2, 4, 8, …. Aha. We get the closed formula \(g_n = 3\cdot 2^{n}\text{.}\)
\((6, 10, 18, 34, 66, \ldots )\text{.}\) 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.
\((15, 33, 57, 87, 123, \ldots)\text{.}\) 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 \(j_n = 3((n+2)(n+3) - 1)\text{.}\)