Item 2.1.1.c.
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. \(a_n = \frac{(n+2)(n+3)}{2} + 2\text{.}\)
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. \(a_n = \frac{(n+2)(n+3)}{2} + 2\text{.}\)