\((3, 6, 10, 15, 21, 28, \ldots)\text{.}\) 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.
in-context