If the sum of the series $$1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \ldots 2 \cdot 6^2 + \ldots$$ upto $$n$$ terms, when $$n$$ is even, is $$\frac{n(n+1)^2}{2}$$, then the sum of the series, when $$n$$ is odd, is

The given series assigns a coefficient $$1$$ to every odd square and a coefficient $$2$$ to every even square.
Hence for the $$k^{\text{th}}$$ term

$$c_k = \begin{cases} 1, & k \text{ odd}\\ 2, & k \text{ even} \end{cases}\,, \qquad T_k = c_k\,k^{2}.$$

Let $$n$$ be even. Write $$n = 2m$$. Splitting the sum into even and odd indices,

$$S_{2m}= \sum_{k=1}^{2m} c_k k^2 = \sum_{\substack{k=1\\k\text{ even}}}^{2m} 2k^2 + \sum_{\substack{k=1\\k\text{ odd}}}^{2m} k^2.$$

Put $$k=2r$$ for even terms and $$k=2r-1$$ for odd terms (with $$r=1,2,\ldots ,m$$):

$$\begin{aligned} S_{2m} &= \sum_{r=1}^{m} 2(2r)^2 + \sum_{r=1}^{m}(2r-1)^2\\ &= 8\sum_{r=1}^{m} r^2 + \sum_{r=1}^{m}\!\bigl(4r^2-4r+1\bigr)\\ &= 12\sum_{r=1}^{m} r^2 - 4\sum_{r=1}^{m} r + m. \end{aligned}$$

Using the standard formulae
$$\sum_{r=1}^{m} r = \frac{m(m+1)}{2}, \qquad \sum_{r=1}^{m} r^2 = \frac{m(m+1)(2m+1)}{6},$$

we get

$$\begin{aligned} S_{2m} &= 12\cdot\frac{m(m+1)(2m+1)}{6} - 4\cdot\frac{m(m+1)}{2} + m\\ &= 2m(m+1)(2m+1) - 2m(m+1) + m\\ &= 2m(m+1)(2m+1 - 1) + m\\ &= 4m^2(m+1) + m\\ &= m(2m+1)^2. \end{aligned}$$

Because $$n=2m$$, this is $$\dfrac{n(n+1)^2}{2}$$, exactly the sum furnished in the question, so our breakdown is consistent.

Now let $$n$$ be odd, say $$n = 2m + 1$$. Then

$$S_{2m+1}=S_{2m} + (2m+1)^2 = m(2m+1)^2 + (2m+1)^2 = (m+1)(2m+1)^2.$$

Substituting $$m = \frac{n-1}{2}$$ and $$2m+1 = n$$ gives

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

Therefore, when $$n$$ is odd the sum of the series equals $$\dfrac{n^{2}(n+1)}{2}$$.

Option C which is: $$\dfrac{n^{2}(n+1)}{2}$$