The sum of first 9 terms of the series $$\frac{1^3}{1} + \frac{1^3 + 2^3}{1+3} + \frac{1^3 + 2^3 + 3^3}{1+3+5} + \ldots$$ is

We begin by observing the pattern of the given series

$$\frac{1^3}{1}\;+\;\frac{1^3+2^3}{1+3}\;+\;\frac{1^3+2^3+3^3}{1+3+5}\;+\;\ldots$$

For the general $$n^{\text{th}}$$ term, the numerator contains the sum of the first $$n$$ cubes and the denominator contains the sum of the first $$n$$ odd numbers. Let us treat each of these sums separately.

Numerator: The sum of cubes formula states

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

Denominator: The sum of the first $$n$$ odd numbers is a well-known result

$$1+3+5+\ldots+(2n-1)=n^2.$$

Hence the $$n^{\text{th}}$$ term $$T_n$$ of the series is

$$T_n=\frac{\displaystyle\left[\dfrac{n(n+1)}{2}\right]^2}{n^2}.$$

We now simplify $$T_n$$. First we expand the square in the numerator:

$$\left[\frac{n(n+1)}{2}\right]^2=\frac{n^2(n+1)^2}{4}.$$

Dividing this by the denominator $$n^2$$ gives

$$T_n=\frac{\,\dfrac{n^2(n+1)^2}{4}\,}{n^2} =\frac{(n+1)^2}{4}.$$

Therefore each term of the series is simply a quarter of a perfect square:

$$T_n=\frac{(n+1)^2}{4}.$$

We are asked for the sum of the first 9 terms, so we must evaluate

$$S_9=\sum_{n=1}^{9}T_n=\sum_{n=1}^{9}\frac{(n+1)^2}{4}.$$

Since the factor $$\frac14$$ is common to every term, we can factor it outside the summation:

$$S_9=\frac14\sum_{n=1}^{9}(n+1)^2.$$

To re-index the summation more comfortably, let us put $$m=n+1$$. Then, when $$n=1$$ we have $$m=2$$ and when $$n=9$$ we have $$m=10$$. Hence

$$\sum_{n=1}^{9}(n+1)^2=\sum_{m=2}^{10}m^2.$$

We now use the formula for the sum of the squares of the first $$k$$ natural numbers:

$$1^2+2^2+\dots+k^2=\frac{k(k+1)(2k+1)}{6}.$$

First we compute the sum up to $$k=10$$:

$$\sum_{m=1}^{10}m^2=\frac{10\,(10+1)\,(2\cdot10+1)}{6} =\frac{10\cdot11\cdot21}{6} =\frac{2310}{6} =385.$$

But we need the sum from $$m=2$$ to $$m=10$$, so we subtract the first term $$1^2=1$$:

$$\sum_{m=2}^{10}m^2=385-1=384.$$

Substituting this result back into our expression for $$S_9$$, we get

$$S_9=\frac14\times384=96.$$

Thus the total of the first nine terms of the given series equals 96.

Hence, the correct answer is Option C.