<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] $$s_1=1^3+2^3+3^3+..n^3=(\frac{n(n+1)}{2})^2\\s_2=1^3+2^3+3^3+..n^3+(n+1)^3+...+(2n)^3=(\frac{2n(2n+1)}{2})^2\\s_2-s_1=(n+1)^3+...+(2n)^3=\\(\frac{2n(2n+1)}{2})^2-(\frac{n(n+1)}{2})^2=\\\frac{n^2}{4}(4(2n+1)^2-(n+1)^2)=\\\frac{n^2}{4}((4n+2+n+1)(4n+2-n-1)$$
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