If $$S_n = 4 + 11 + 21 + 34 + 50 + \ldots$$ to $$n$$ terms, then $$\frac{1}{60}S_{29} - S_9$$ is equal to

Consider the series $$S_n = 4 + 11 + 21 + 34 + 50 + \ldots$$. The differences between successive terms are $$7, 10, 13, 16, \ldots$$, which form an arithmetic progression with first term 7 and common difference 3.

To determine the $$n$$-th term, observe that starting from 4, each subsequent increase is of the form $$3k + 4$$ for $$k = 1,2,\dots,n-1$$. Thus,

$$a_n = 4 + \sum_{k=1}^{n-1}(3k + 4) = 4 + \frac{3(n-1)n}{2} + 4(n-1) = \frac{n(3n+5)}{2}$$.

It is straightforward to verify that $$a_1 = 4$$, $$a_2 = 11$$, and $$a_3 = 21$$.

Summing these terms up to $$n$$ gives

$$S_n = \sum_{k=1}^{n} \frac{k(3k+5)}{2} = \frac{1}{2}\Bigl[3\cdot\frac{n(n+1)(2n+1)}{6} + 5\cdot\frac{n(n+1)}{2}\Bigr]$$
$$= \frac{n(n+1)}{4}(2n + 1 + 5) = \frac{n(n+1)(n+3)}{2}$$.

In particular,

$$S_{29} = \frac{29 \times 30 \times 32}{2} = 13920,$$
$$S_9 = \frac{9 \times 10 \times 12}{2} = 540.$$

Hence, the required value is

$$\frac{1}{60}\bigl(S_{29} - S_9\bigr) = \frac{13920 - 540}{60} = \frac{13380}{60} = 223.$$