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.
 A: Consider the sequence of consecutive integers beginning with one more than the $(n-1)$th triangular number and ending with $(n-1)$ more than the first term. This sequence has $n$ terms. If you take the terms of this sequence, cube them, and add them together, you get the $n$th term of the sequence that you want, which you can call $T_n$.
The $(n-1)$th triangular number is $\frac{n^2-n}2$. One more than that is $\frac{n^2-n+2}{2}$. And $(n-1)$ more than the latter is $\frac{n^2-n+2}{2} + (n-1) = \frac{n^2+n}{2}$.
Therefore $T_n$ can be computed as the difference between $C(\frac{n^2+n}{2})$ and $C(\frac{n^2-n}{2})$, where $C(m)$ represents the sum of consecutive cubes from $1$ to $m$, i.e.
$$T_n = \sum_{\frac{n^2-n}{2}}^{\frac{n^2+n}{2}}k^3 = C(\frac{n^2+n}{2}) - C(\frac{n^2-n}{2})$$
Now $C(m) = \frac 14m^2(m+1)^2$ (the sum of the cubes of the first $m$ positive integers is the square of the sum of the first $m$ positive integers, which is also the square of the $m$th triangular number - a standard result which is a particular case of the general Faulhaber's formula).
Hence,
$$\begin{align}T_n &= \frac 14 \left(\left(\frac{n^2+n}{2}\right)^2\left(\frac{n^2+n}{2}+1\right)^2 - \left(\frac{n^2-n}{2}\right)^2\left(\frac{n^2-n}{2}+1\right)^2 \right) \\&= \frac 18n^3(n^2+1)(n^2+3)\end{align}$$
The final simplification is tedious by hand, but an online computer algebra system makes it trivial. You can verify the formula works for the first few terms, e.g. $T_1 = 1, T_2 = 35, T_3 = 405$ and so forth.
A: Let me see if I understand you correctly for I am a little bit confused about your notation. I guess that by the first term of the sequence you mean $1^3$ and $(2^3 + 3^3)$ is the second term, $(4^3 + 5^3 + 6^3)$ is the third term and so on. If this is so, it seems correct that the firs summand of the $n$-th term is given by 
$$(\frac{n(n-1)}{2}+1)^3 \ .$$ 
Since the $n$-th term contains $n$ summands, the last summand of the $n$-th term is given by 
$$(\frac{n(n-1)}{2}+1 + (n-1))^3 = (\frac{n(n+1)}{2})^3 \ , $$
where we added $(n-1)$ because the first summand is to be raised $(n-1)$ times by $1$ to get the last summand of the $n$-th term (your formula is almost the same for it seems that you used the same logic and only forgot to add $+1$). This expression will serve as a checker after we express the general formula for the $n$-th term.
To describe the $n$-th term should be straightforward now for we know the first summand and the number of steps we have to make to get to the last summand for arbitrary $n \in \mathbb{N}$. The $n$-th term should be 
$$ \sum_{i=0}^{n-1} (\frac{n(n-1)}{2}+1 +i)^3 \ . $$ 
The first summand is obviously in accordance with what we know and if we compute the last summand of the above expression we get
$$ (\frac{n(n-1)}{2}+1 + n-1)^3 = (\frac{n(n-1)}{2} + n)^3 = (\frac{n(n+1)}{2} )^3 \ ,$$ 
which is what we expected. 
In case this answer is not what you were looking for, try to specify more carefully what you mean by $n$-th term.
A: 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 
$$1+2+...+n = \dfrac{n(n+1)}{2}$$
$$1+2+...+(n-1)=\dfrac{n(n-1)}{2}.$$
A: 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
