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1. Note that if we subtract 1 from each term, we get the square numbers. Thus \(a_n = n^2 + 1\text{.}\)

2. These look like the triangular numbers, only shifted by 1. We get: \(a_n = \frac{n(n+1)}{2} - 1\text{.}\)

3. If you subtract 2 from each term, you get triangular numbers, only starting with 6 instead of 1. So we must shift vertically and horizontally. \(a_n = \frac{(n+2)(n+3)}{2} + 2\text{.}\)

4. These seem to grow very quickly. Further, if we add 1 to each term, we find the factorials, although starting with 2 instead of 1. This gives, \(a_n = (n+1)! - 1\) (where \(n! = 1 \cdot 2 \cdot 3 \cdots n\)).