<s>[INST] Formulae for sequences Given that for $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$
deduce that $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3 = \frac{n^2(3n+1)(5n+3)}{4}$
So far:
the sequence $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3$ gives $2^3 + 3^3 + 4^3 +\cdots,$ when n=1.
The brackets in the formula for the second sequence are $2n$ and $4n+2$ bigger than $(n+1)$ in the original i.e. $(3n+1)(5n+3).$
 [/INST] You can use the following relation :
$$\sum_{k=n+1}^{2n}k^3=\sum_{k=1}^{\color{red}{2n}}k^3-\sum_{k=1}^{n}k^3$$
(To get the first sum on RHS, you only need to replace the $n$ in the given equation with $\color{red}{2n}$.)
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