Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the first element of every individual term makes a sequence $1, 2, 4, 7...$.  $\therefore$ the first element of each term is $(1+\frac{n(n-1)}2)^3$ and this the last element is $(\frac{n(n-1)}2 + 1 + (n-1))^3$. But I don't understand what to do next.

HINT
Note that $$a_n= \sum_1^{T(n)} k^3-\sum_1^{T(n-1)} k^3$$
where $T(n)=\frac{n(n+1)}{2}$ are triangular numbers
