It is given that $f(r)=(r-2)2^{r-1}$. Show that $f(r+1)-f(r)=2^r a r$, where $a$ is a constant to be determined. Hence find $\sum_{r=1}^n r2^r$. Using the result in part (ii), find $\sum_{r=3}^n (r+2)2^r$ in terms of n.

1. Substitute $r+1$ into $f(r)$ and write the expression for $f(r+1) - f(r)$.

$f(r+1) = (r+1-2)2^{r+1-1} = (r-1)2^r$

$f(r+1) - f(r) = (r-1)2^r - (r-2)2^{r-1}$

2. Factor out $2^r$ and simplify the expression to find the value of $a$.

$f(r+1) - f(r) = (r-1)2^r - \frac{1}{2}(r-2)2^r$
$= 2^r \left( r - 1 - \frac{1}{2}r + 1 \right)$
$= 2^r \left( \frac{1}{2}r \right)$

Comparing this with $2^r a r$, we get $a = \frac{1}{2}$.

1. Determine the constant $a$ from part (i) to express the summand $r2^r$ in terms of $f(r+1) - f(r)$.

$f(r+1) - f(r) = (r+1-2)2^{r+1-1} - (r-2)2^{r-1}$
$= (r-1)2^r - (r-2)\frac{1}{2}2^r$
$= 2^r \left( r - 1 - \frac{1}{2}r + 1 \right) = \frac{1}{2} r 2^r$
$\therefore a = \frac{1}{2} \implies r 2^r = 2(f(r+1) - f(r))$

2. Set up the summation using the relationship found and write out the terms to show the telescoping cancellation.

$\sum_{r=1}^n r 2^r = \sum_{r=1}^n 2(f(r+1) - f(r))$
$= 2 \big[ (f(2) - f(1)) + (f(3) - f(2)) + \dots + (f(n+1) - f(n)) \big]$

3. Evaluate the surviving terms to find the final expression for the sum.

$= 2 \big[ f(n+1) - f(1) \big]$
Since $f(n+1) = (n-1)2^n$ and $f(1) = (1-2)2^{1-1} = -1$:
$\text{Sum} = 2 \big[ (n-1)2^n - (-1) \big]$
$= 2 \big[ (n-1)2^n + 1 \big]$
$= (n-1)2^{n+1} + 2$

1. Concisely derive the result of part (ii) as it is needed to solve the current part.

From part (i), $f(r+1) - f(r) = (r-1)2^r - (r-2)2^{r-1} = \frac{1}{2} r 2^r$.
Thus, $\sum_{r=1}^n r 2^r = 2 \sum_{r=1}^n (f(r+1) - f(r))$.
Using the method of differences:
$\sum_{r=1}^n r 2^r = 2 \left[ f(n+1) - f(1) \right] = 2 \left[ (n-1)2^n - (-1) \right] = (n-1)2^{n+1} + 2$.

2. Split the target sum and evaluate the first component by adjusting the indices of the result from part (ii).

The target sum can be split into two parts:
$\sum_{r=3}^n (r+2)2^r = \sum_{r=3}^n r 2^r + 2 \sum_{r=3}^n 2^r$

To find $\sum_{r=3}^n r 2^r$, we subtract the first two terms from the result of part (ii):
$\sum_{r=3}^n r 2^r = \sum_{r=1}^n r 2^r - (1 \cdot 2^1 + 2 \cdot 2^2)$
$= \left[ (n-1)2^{n+1} + 2 \right] - (2 + 8)$
$= (n-1)2^{n+1} - 8$

3. Evaluate the second component, which is a geometric series.

The term $2 \sum_{r=3}^n 2^r$ involves a geometric series with first term $a = 2^3 = 8$, common ratio $R = 2$, and $n - 3 + 1 = n - 2$ terms:
$2 \sum_{r=3}^n 2^r = 2 \left[ \frac{8(2^{n-2} - 1)}{2 - 1} \right]$
$= 16(2^{n-2} - 1)$
$= 2^4 \cdot 2^{n-2} - 16$
$= 2^{n+2} - 16$

4. Combine the components and simplify to find the final expression.

Substitute the evaluated components back into the split sum:
$\sum_{r=3}^n (r+2)2^r = \left[ (n-1)2^{n+1} - 8 \right] + \left[ 2^{n+2} - 16 \right]$
$= (n-1)2^{n+1} + 2 \cdot 2^{n+1} - 24$
$= (n-1+2)2^{n+1} - 24$
$= (n+1)2^{n+1} - 24$