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}

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