What is the sum of the first 28 terms in the following sequence? 1, 1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, …

• The sequence is composed of blocks where the $n$-th block starts with $n$ and decreases by 1 until it reaches 1, so the $n$-th block is $(n, n-1, \dots, 1)$.

• The length of the $n$-th block is $n$ terms, and the sum of the $n$-th block is $\frac{n(n+1)}{2}$.

• We need the sum of the first 28 terms. The sum of the lengths of the first $k$ blocks is $\frac{k(k+1)}{2}$.

• For $k=6$, the total number of terms is $\frac{6(7)}{2} = 21$. For $k=7$, the total number of terms is $\frac{7(8)}{2} = 28$.

• Thus, the first 28 terms exactly comprise the first 7 blocks.

• The total sum is the sum of the sums of the first 7 blocks: $\sum_{n=1}^{7} \frac{n(n+1)}{2} = \frac{1}{2} \sum_{n=1}^{7} (n^2 + n) = \frac{1}{2} (\frac{7(8)(15)}{6} + \frac{7(8)}{2}) = \frac{1}{2} (140 + 28) = \frac{168}{2} = 84$.