If the sum of the first 15 terms of the series $$\left(\frac{3}{4}\right)^3 + \left(1\frac{1}{2}\right)^3 + \left(2\frac{1}{4}\right)^3 + 3^3 + \left(3\frac{3}{4}\right)^3 + \ldots$$ is equal to 225K, then K is equal to:

The given series begins as

$$\left(\frac34\right)^3 + \left(1\frac12\right)^3 + \left(2\frac14\right)^3 + 3^3 + \left(3\frac34\right)^3 + \ldots$$

First we write every mixed fraction as an ordinary fraction or decimal so that the pattern is clear:

$$\frac34, \; 1\frac12 = \frac32, \; 2\frac14 = \frac94, \; 3 = 3, \; 3\frac34 = \frac{15}4, \ldots$$

Numerically these are

$$0.75,\;1.5,\;2.25,\;3,\;3.75,\; \ldots$$

We observe that each successive base increases by $$0.75=\frac34$$. Hence the un-cubed terms form an arithmetic progression with

first term $$a = 0.75=\frac34,$$

common difference $$d = 0.75=\frac34.$$

Therefore the $$n^{\text{th}}$$ base before cubing is obtained from the standard A.P. formula $$a_n = a + (n-1)d.$$ Substituting $$a=\frac34$$ and $$d=\frac34,$$ we get

$$a_n = \frac34 + (n-1)\frac34 = \frac34\,n.$$

Each actual term of the series is the cube of this base, so the $$n^{\text{th}}$$ term $$T_n$$ equals

$$T_n = \left(\frac34\,n\right)^3 = \frac{27}{64}\,n^3.$$

We require the sum of the first 15 terms, i.e.

$$S_{15} = \sum_{n=1}^{15} T_n = \sum_{n=1}^{15} \frac{27}{64}\,n^3.$$

The constant factor $$\frac{27}{64}$$ can be taken outside the summation:

$$S_{15} = \frac{27}{64}\,\sum_{n=1}^{15} n^3.$$

Now we use the well-known formula for the sum of the cubes of the first $$N$$ natural numbers, stated as

$$\sum_{n=1}^{N} n^3 = \left[\frac{N(N+1)}{2}\right]^2.$$

Putting $$N = 15$$ we obtain

$$\sum_{n=1}^{15} n^3 = \left[\frac{15\cdot16}{2}\right]^2 = (120)^2 = 14400.$$

Substituting this value into the expression for $$S_{15}$$ gives

$$S_{15} = \frac{27}{64}\times14400.$$

We simplify step by step. First divide $$14400$$ by $$64$$:

$$\frac{14400}{64} = 225.$$

So

$$S_{15} = 27 \times 225.$$

Next multiply:

$$27 \times 225 = 27 \times (200 + 25) = 27\cdot200 + 27\cdot25 = 5400 + 675 = 6075.$$

Therefore

$$S_{15} = 6075.$$

According to the question, this sum equals $$225K$$, that is

$$225K = 6075.$$

Solving for $$K$$ by dividing both sides by $$225$$, we have

$$K = \frac{6075}{225}.$$

Since $$225 \times 27 = 6075$$, the division yields

$$K = 27.$$

Hence, the correct answer is Option B.