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

(a)
Let P(n) be the statement \sum_{r=1}^n r(r+2)(r+5) = \frac{1}{12}n(n+1)(3n^2+31n+74) \forall n \in \mathbb{Z^+}

When n=1, LHS = 1 \times 3 \times 6 = 18
RHS = \frac{1}{12}\times 1 \times 2 \times (3+31+74) = 18

Since LHS = RHS, P(1) is true.

Assume P(k) is true for some k, k \in \mathbb{Z^+}

\sum_{r=1}^k r(r+2)(r+5) = \frac{1}{12}k(k+1)(3k^2+31k+74)

Want to show P(k+1) is true, \sum_{r=1}^{k+1} r(r+2)(r+5) = \frac{1}{12}(k+1)(k+2)(3k^3+43k^2+182k+216)

LHS
= \sum_{r=1}^{k+1} r(r+2)(r+5)

= \frac{1}{12}k(k+1)(3k^2+31k+74) + (k+1)(k+3)(k=6)

= \frac{1}{12}(k+1)[3k^3 + 321k^2 + 74k + 12(k^2 + 9k +18)]

= \frac{1}{12}(k+1)(3k^3+43k^2+182k+216)

= RHS

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

(b)
(i)
A = 1, B = -1

(ii)
\sum_{r=1}^n \frac{1}{2r+1} - \frac{1}{2r+3}

= \frac{1}{3} - \frac{1}{5}

+ \frac{1}{5} - \frac{1}{7}

\ldots

+ \frac{1}{2n-1} - \frac{1}{2n+1}

+ \frac{1}{2n+1} - \frac{1}{2n+3}

= \frac{1}{3} - \frac{1}{2n+3}

(iii)

S_{\infty} = \frac{1}{3}

S_n - S_{\infty} = \frac{1}{2n+3} ~ \textless ~10^{-3}

Least n = 499

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