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]

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

Hence or otherwise find $\sum\limits_{r=1}^{n}{{{\left( 2r-1 \right)}^{3}}}$, simplifying your answer.

Recurrence Relations and Summation

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2006 P1 Q11 [Modified]

**Given:** $\displaystyle\sum_{r=1}^{n} r^3 = 1^3 + 2^3 + ... + n^3 = \frac{1}{4}n^2(n+1)^2$

**Part 1: Prove** $2^3 + 4^3 + 6^3 + ... + (2n)^3 = 2n^2(n+1)^2$

LHS $= 2^3 + 4^3 + 6^3 + ... + (2n)^3$

$= \displaystyle\sum_{r=1}^{n} (2r)^3$

$= \displaystyle\sum_{r=1}^{n} (2^3 r^3)$

$= 2^3 \displaystyle\sum_{r=1}^{n} r^3$

$= 8 \left[\frac{1}{4}n^2(n+1)^2\right]$

$= 2n^2(n+1)^2 = $ RHS (shown)

**Part 2: Find** $\displaystyle\sum_{r=1}^{n} (2r-1)^3$

Note:
$\displaystyle\sum_{r=1}^{n} (2r-1)^3 = 1^3 + 3^3 + 5^3 + ... + (2n-1)^3$

$\displaystyle\sum_{r=1}^{2n} r^3 = 1^3 + 2^3 + 3^3 + 4^3 + ... + (2n-1)^3 + (2n)^3$ ... (1)

$\displaystyle\sum_{r=1}^{n} (2r)^3 = 2^3 + 4^3 + ... + (2n)^3$ ... (2)

(1) - (2):

$\displaystyle\sum_{r=1}^{2n} r^3 - \sum_{r=1}^{n} (2r)^3 = \sum_{r=1}^{n} (2r-1)^3$

$\therefore \displaystyle\sum_{r=1}^{n} (2r-1)^3 = \sum_{r=1}^{2n} r^3 - \sum_{r=1}^{n} (2r)^3$

$= \frac{1}{4}(2n)^2(2n+1)^2 - 2n^2(n+1)^2$

$= n^2(2n+1)^2 - 2n^2(n+1)^2$

$= n^2\left[(2n+1)^2 - 2(n+1)^2\right]$

$= n^2\left[4n^2 + 4n + 1 - 2(n^2 + 2n + 1)\right]$

$= n^2\left[4n^2 + 4n + 1 - 2n^2 - 4n - 2\right]$

$= n^2(2n^2 - 1)$

$= 2n^4 - n^2$

N2006 P1 Q11 [Modified]

N2006 P1 Q11 [Modified]

Video Solution: