Find $\sum_{r=0}^{n} [(n+2)r + n]$, giving your answer in terms of $n$. [You may use the result $\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2$ for the rest of this question.] By writing down the first two and the last two terms in the series, find $\sum_{r=1}^{n} (r+2)^3$, giving your answer in terms of $n$. Find $1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + \ldots + (2n-1)^3 - (2n)^3$, giving your answer in terms of $n$.

1. Split the sum into terms involving $r$ and constant terms.

$\sum_{r=0}^{n} [(n+2)r + n] = (n+2)\sum_{r=0}^{n} r + \sum_{r=0}^{n} n$

2. Evaluate the individual sums, noting that there are $n+1$ terms from $r=0$ to $r=n$.

$\sum_{r=0}^{n} r = \frac{1}{2}n(n+1)$
$\sum_{r=0}^{n} n = n(n+1)$

3. Substitute the evaluated sums back and factorise completely.

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

1. Write down the first two and last two terms of the series as instructed.

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

2. Express the series as a difference involving a standard sum starting from $r=1$.

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

3. Substitute the given formula for the sum of cubes and simplify.

Using $\sum_{r=1}^{N} r^3 = \frac{1}{4}N^2(N+1)^2$ with $N = n+2$:
$= \frac{1}{4}(n+2)^2((n+2)+1)^2 - (1 + 8)$
$= \frac{1}{4}(n+2)^2(n+3)^2 - 9$

1. Express the alternating series in terms of the sum of all cubes and the sum of even cubes.

Let the required sum be $S$.
The series alternates signs, which can be evaluated by taking the sum of all positive cubes up to $2n$, and subtracting twice the sum of the even cubes.
$S = \sum_{r=1}^{2n} r^3 - 2 \sum_{r=1}^{n} (2r)^3$

2. Evaluate both summation terms using the given standard formula.

Substitute $2n$ into the given result $\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2$ for the first sum:
$\sum_{r=1}^{2n} r^3 = \frac{1}{4}(2n)^2(2n+1)^2 = n^2(2n+1)^2$

For the second sum, extract the constant factor $2^3$ first:
$2 \sum_{r=1}^{n} (2r)^3 = 2 \sum_{r=1}^{n} 8r^3 = 16 \sum_{r=1}^{n} r^3$
$= 16 \left[ \frac{1}{4}n^2(n+1)^2 \right] = 4n^2(n+1)^2$

3. Substitute the evaluated sums back into $S$ and simplify the expression.

$S = n^2(2n+1)^2 - 4n^2(n+1)^2$
Factorise by taking out the common factor $n^2$:
$S = n^2 \left[ (2n+1)^2 - 4(n+1)^2 \right]$
Using the difference of two squares on the expression in the bracket:
$S = n^2 \left[ (2n+1) - 2(n+1) \right] \left[ (2n+1) + 2(n+1) \right]$
$S = n^2 (2n + 1 - 2n - 2)(2n + 1 + 2n + 2)$
$S = n^2 (-1)(4n + 3)$
$S = -n^2(4n+3)$