No you cannot. You might guess that the next terms are:

1. \(7\)
2. \(-3\)
3. \(4\)

4. 64

5. 49

6. 34

7. 28

8. 17

9. \(-2\)
10. \(24\)

In fact, those are the next terms of the sequences I had in mind when I made up the example, but there is no way to be sure they are correct.

Still, we will often do this. Given the first few terms of a sequence, we can ask what the pattern in the sequence suggests the next terms are.

We know that \(a_6 = 2a_5 - a_4\). So to find \(a_6\) we need to find \(a_5\) and \(a_4\). Well \begin{equation*} a_5 = 2a_4 - a_3 \qquad \text{and} \qquad a_4 = 2a_3 - a_2, \end{equation*} so if we can only find \(a_3\) and \(a_2\) we would be set. Of course \begin{equation*} a_3 = 2a_2 - a_1 \qquad \text{and} \qquad a_2 = 2a_1 - a_0, \end{equation*} so we only need to find \(a_1\) and \(a_0\). But we are given these. Thus \begin{align*} a_0 \amp = 3\\ a_1 \amp = 4\\ a_2 \amp = 2\cdot 4 - 3 = 5\\ a_3 \amp = 2\cdot 5 - 4 = 6\\ a_4 \amp = 2\cdot 6 - 5 = 7\\ a_5 \amp = 2\cdot 7 - 6 = 8\\ a_6 \amp = 2\cdot 8 - 7 = 9. \end{align*}

Note that now we can guess a closed formula for the \(n\)th term of the sequence: \(a_n = n+3\). To be sure this will always work, we could plug in this formula into the recurrence relation: \begin{align*} 2a_{n-1} - a_{n-2} \amp = 2((n-1) + 3) - ((n-2) + 3)\\ \amp = 2n + 4 - n - 1 \\ \amp = n + 3\\ \amp = a_n. \end{align*}

That is not quite enough though, since there can be multiple closed formulas that satisfy the same recurrence relation; we must also check that our closed formula agrees on the initial terms of the sequence. Since \(a_0 = 0 + 3 = 3\) and \(a_1 = 1+3 = 4\) are the correct initial conditions, we can now conclude we have the correct closed formula.

Before you say this is impossible, what we are asking for is simply to find a closed formula which agrees with all of the initial terms of the sequences. Of course there is no way to read into the mind of the person who wrote the numbers down, but we can at least do this.

Now the first few terms of \((T_n)_{n\ge 0}\) are \(0, 1, 3, 6, 10, 15, 21, \ldots\) (these are called the triangular numbers). The first few terms of \((a_n)_{n\ge 0}\) are \(1, 2, 4, 8, 16, \ldots\). Now let's try to find formulas for the given sequences.

1. \((1, 2, 4, 7, 11, 16, 22, \ldots)\). Note that if subtract 1 from each term, we get the sequence \((T_n)\). So we have \(b_n = T_n + 1\). Therefore a closed formula is \(b_n = \frac{n(n+1)}{2} + 1\). A quick check of the first few \(n\) confirms we have it right.

2. \((3, 5, 9, 17, 33, \ldots )\). Each term in this sequence is one more than a power of 2, so we might guess the closed formula is \(c_n = a_n+1 = 2^n + 1\). If we try this though, we get \(c_0 2^0 + 1 = 2\) and \(c_1 = 2^1 + 1 = 3\). We are off because the indices are shifted. What we really want is \(c_n = a_{n+1}+1\) giving \(c_n = 2^{n+1} + 1\).

3. (\(0, 2, 6, 12, 20, 30, 42,\ldots \)). Notice that all these terms are even. What happens if we factor out a 2? We get \((T_n)\)! More precisely, we find that \(d_n/2 = T_n\), so this sequence has closed formula \(d_n = n(n+1)\).

4. \((3, 6, 10, 15, 21, 28, \ldots)\). 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 \(T_n\), 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.

5. \((0, 1, 3, 7, 15, 31, \ldots )\). Try adding one to each term and we get powers of 2. You might guess this because each term is approximately twice the previous term. Closed formula: \(f_n = 2^{n} - 1\).

6. \((3, 6, 12, 24, 48, \ldots )\). These numbers are also doublig each time, but are also all multiples of 3. Dividing each by 3 gives 1, 2, 4, 8, …. Aha. We get the closed formula \(g_n = 3\cdot 2^{n}\).

7. \((6, 10, 18, 34, 66, \ldots )\). To get from one term to the next, we almost double each term. So maybe we can relate this back to \(2^n\). 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 \(2^2\), 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 \(c_n = 2^{n+1} + 1\). Our sequence here would be twice this, so \(h_n = 2(2^n + 1)\), which is the same as we got before.

8. \((15, 33, 57, 87, 123, \ldots)\). Try dividing each term by 3. That gives the sequence \(5, 11, 19, 29, 41,\ldots\). Now add one: \(6, 12, 20, 30, 42, \ldots\), which is \((d_n)\) in this example, except starting with 6 instead of 0. So let's start with the formula \(d_n= n(n+1)\). To start with the 6, we shift: \((n+2)(n+3)\). But this is one too many, so subtract 1: \((n+2)(n+3) - 1\). That gives us our sequence, but divided by 3. So we want \(j_n = 3((n+2)(n+3) - 1)\).