All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let $P(n)$ be the statement: $\sum_{r=1}^n r(r+2) = \frac{1}{6}n(n+1)(2n+7), n \in \mathbb{Z}^+$

When $n = 1, ~\text{LHS} = 1(3)=3, \text{~and~RHS} = \frac{1}{6}(2)(9) = 3 = \text{LHS}$

$\therefore, P(n)$ is true.

Assume that $P(k)$ is true for some $k \in \mathbb{Z}^+$, i.e. $\sum_{r=1}^k r(r+2) = \frac{1}{6}k(k+1)(2k+7)$

Want to prove that $P(k+1)$ is true, i.e. $\sum_{r=1}^{k+1} r(r+2) = \frac{1}{6}(k+1)(k+2)(2k+9)$

LHS

$= \sum_{r=1}^{k+1} r(r+2)$

$= \sum_{r=1}^{k} r(r+2) + (k+1)(k+3)$

$= \frac{1}{6}k(k+1)(2k+7) + (k+1)(k+3)$

$= \frac{1}{6}(k+1)[k(2k+7)+6(k+3)]$

$= \frac{1}{6}(k+1)(2k^2+13k+18)$

$= \frac{1}{6}(k+1)(k+2)(2k+9)$

$= \text{RHS}$

Since $P(1)$ is true and $P(k) \text{~is~true} \Rightarrow P(k+1)$ is true, hence by Mathematical Induction, $P(n)$ is true for all $n \in \mathbb{Z}^+$

(ii)
(a)

$\frac{1}{r(r+2)} = \frac{1}{2r} - \frac{1}{2(r+2)}$

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

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

$= \frac{1}{2} - \frac{1}{2(3)}$

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

$+ \frac{1}{2(3)} - \frac{1}{2(5)}$

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

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

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

$= \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}$

(b)
As $n \rightarrow \infty, \frac{1}{2(n+1)} \rightarrow 0, \frac{1}{2(n+2)} \rightarrow 0$

Hence, $\sum_{r=1}^{\infty} \frac{1}{r(r+2)}$ is a convergent series, and the value of the sum to infinity is $\frac{3}{4}$