<s>[INST] 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.
 [/INST] Hint: The sum is equal to
$$S_n=1^3+2^3+3^3+...+n^3=\dfrac{n^2(n+1)^2}{4}$$
The $n^\text{th}$ term in your representation is given as 
$$S_{n(n+1)/2}-S_{n(n-1)/2}.$$
The indices are just the partial sums for 
$$1+2+...+n = \dfrac{n(n+1)}{2}$$
$$1+2+...+(n-1)=\dfrac{n(n-1)}{2}.$$
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