The sum of the series : $$(2)^2 + 2(4)^2 + 3(6)^2 + \ldots$$ upto 10 terms is :

The given series is $$(2)^2 + 2(4)^2 + 3(6)^2 + \ldots$$ upto 10 terms. To find the sum, we first identify the general term of the series. Observing the pattern, the coefficients are natural numbers $$1, 2, 3, \ldots$$, and the bases are even numbers $$2, 4, 6, \ldots$$. The $$n$$th even number is $$2n$$. Therefore, the $$n$$th term of the series is given by $$T_n = n \times (2n)^2$$.

Simplifying the expression: $$T_n = n \times (2n)^2 = n \times 4n^2 = 4n^3$$. So, $$T_n = 4n^3$$.

The sum of the first 10 terms is $$S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} 4n^3$$. We can factor out the constant 4: $$S = 4 \times \sum_{n=1}^{10} n^3$$.

The sum of cubes of the first $$k$$ natural numbers is given by $$\sum_{n=1}^{k} n^3 = \left[ \frac{k(k+1)}{2} \right]^2$$. Substituting $$k = 10$$: $$\sum_{n=1}^{10} n^3 = \left[ \frac{10 \times 11}{2} \right]^2 = \left[ \frac{110}{2} \right]^2 = [55]^2 = 3025$$.

Therefore, $$S = 4 \times 3025 = 12100$$.

Hence, the sum of the series upto 10 terms is 12100. Comparing with the options, 12100 corresponds to option C.

Hence, the correct answer is Option C.