A sequence $u_0, u_1, u_2, \dots$ is such that $u_0 = 1$ and $u_n = u_{n-1} + 2n + \frac{1}{n(n+1)}, \quad\text{for all } n \in \mathbb{N}^+.$

1. Express the rational function in partial fractions.

Let $\dfrac{1}{r(r+1)} = \dfrac{A}{r} + \dfrac{B}{r+1}$.

$1 = A(r+1) + Br$

Substituting $r = 0$, we get $A = 1$.
Substituting $r = -1$, we get $B = -1$.

Therefore, $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$.

2. Split the summation and evaluate the arithmetic sequence portion.

$\displaystyle \sum_{r=1}^n\left(2r + \dfrac{1}{r(r+1)}\right) = \sum_{r=1}^n 2r + \sum_{r=1}^n \left(\dfrac{1}{r} - \dfrac{1}{r+1}\right)$

For the first sum, it is an arithmetic progression. Using the sum formula for an AP or standard sum formula:
$\displaystyle \sum_{r=1}^n 2r = 2 \left( \dfrac{n(n+1)}{2} \right) = n(n+1)$

3. Evaluate the second summation using the method of differences.

$\displaystyle \sum_{r=1}^n \left(\dfrac{1}{r} - \dfrac{1}{r+1}\right) = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \dots + \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right)$

Since intermediate terms cancel out in pairs, we are left with the first and last terms:
$= 1 - \dfrac{1}{n+1}$

4. Combine the evaluated parts to reach the final expression.

Total sum $= n(n+1) + 1 - \dfrac{1}{n+1}$

1. Concisely obtain the sum evaluated in part (i), as it is required for the target.

From part (i), we use the partial fractions $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ to evaluate the sum:
$\sum_{r=1}^n\left(2r + \dfrac{1}{r(r+1)}\right) = \sum_{r=1}^n 2r + \sum_{r=1}^n \left(\dfrac{1}{r} - \dfrac{1}{r+1}\right)$
$= 2\left(\dfrac{n(n+1)}{2}\right) + \left( (1 - \dfrac{1}{2}) + (\dfrac{1}{2} - \dfrac{1}{3}) + \dots + (\dfrac{1}{n} - \dfrac{1}{n+1}) \right)$
$= n(n+1) + 1 - \dfrac{1}{n+1}$

2. Rearrange the given recurrence relation to express $u_r - u_{r-1}$ and sum both sides from $r=1$ to $n$.

From the given recurrence relation, replace $n$ with $r$ and rearrange:
$u_r = u_{r-1} + 2r + \dfrac{1}{r(r+1)}$
$u_r - u_{r-1} = 2r + \dfrac{1}{r(r+1)}$
Taking the sum from $r=1$ to $n$ on both sides:
$\sum_{r=1}^n (u_r - u_{r-1}) = \sum_{r=1}^n \left(2r + \dfrac{1}{r(r+1)}\right)$

3. Evaluate the left-hand side using the Method of Differences.

$\sum_{r=1}^n (u_r - u_{r-1}) = (u_1 - u_0) + (u_2 - u_1) + \dots + (u_n - u_{n-1})$
Notice that all intermediate terms cancel out, leaving only the first and last terms:
$\sum_{r=1}^n (u_r - u_{r-1}) = u_n - u_0$

4. Substitute the given value of $u_0$ and the evaluated sum from part (i) to complete the proof.

Equating the evaluated left side and right side sums:
$u_n - u_0 = n(n+1) + 1 - \dfrac{1}{n+1}$
Substitute the given initial condition $u_0 = 1$:
$u_n - 1 = n(n+1) + 1 - \dfrac{1}{n+1}$
$u_n = n(n+1) + 2 - \dfrac{1}{n+1} \quad \text{(shown)}$

1. Evaluate the result of the sum from part (i) as it is required for this part.

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

2. Rewrite the target sum by shifting the index to relate it to the sum in part (i).

Let $k = r + 1$. When $r=1$, $k=2$; when $r=n-1$, $k=n$.
$\sum_{r=1}^{n-1}\left(2(r+1) + \frac{1}{(r+1)(r+2)}\right) = \sum_{k=2}^n \left(2k + \frac{1}{k(k+1)}\right)$
$= \sum_{k=1}^n \left(2k + \frac{1}{k(k+1)}\right) - \left( 2(1) + \frac{1}{1(2)} \right)$

3. Substitute the evaluated sum from part (i) and simplify.

$= \left( n(n+1) + 1 - \frac{1}{n+1} \right) - \left( 2 + \frac{1}{2} \right)$
$= n^2 + n + 1 - \frac{1}{n+1} - \frac{5}{2}$
$= n^2 + n - \frac{3}{2} - \frac{1}{n+1}$