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