Find the sum of the following series up to $n$ terms:
$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$

The $n^{th}$ term of the given series is
$a_n = \frac{1^{3}+2^{3}+3^{3}+\ldots+n^{3}}{1+3+5+\ldots+(2 n-1)} = \frac{[\frac{n(n+1)}{2}]^{2}}{n^2}$
Since $1+3+5+\ldots+(2 n-1)$ is an $A.P.$ with $n$ terms,its sum is $n^2$.
$a_n = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4} = \frac{1}{4}(n^2 + 2n + 1)$
Now,the sum of $n$ terms $S_n = \sum_{k=1}^n a_k = \frac{1}{4} \sum_{k=1}^n (k^2 + 2k + 1)$
$S_n = \frac{1}{4} [\frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} + n]$
$S_n = \frac{1}{4} [\frac{n(n+1)(2n+1) + 6n(n+1) + 6n}{6}]$
$S_n = \frac{n}{24} [2n^2 + 3n + 1 + 6n + 6 + 6] = \frac{n(2n^2 + 9n + 13)}{24}$