Given $\sum_{r=1}^{n} r^3 = \frac{1}{4} n^2 (n+1)^2$ show that $2^3 + 4^3 + 6^3 + \dots + (2n)^3 = 2

Question

Given $\sum_{r=1}^{n} r^3 = \frac{1}{4} n^2 (n+1)^2$ show that $2^3 + 4^3 + 6^3 + \dots + (2n)^3 = 2n^2 (n+1)^2$. [3] Hence or otherwise find $\sum\limits_{r=1}^{n}{{{\left( 2r-1 \right)}^{3}}}$, simplifying your answer. [3]

Accepted Answer

(i) $2^3+4^3+\cdots+(2n)^3=2n^2(n+1)^2$ (ii) $\sum_{r=1}^n(2r-1)^3=n^2(2n^2-1)$

N2006 P1 Q11 [Modified]

N2006 P1 Q11 [Modified]

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