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 preposition $\sum_{r=1}^{n} r(2r^{2}+1) = \frac{n}{2} (n+1)(n^{2}+n+1), n \in {\mathbb{Z}}^{+}.$

When $n=1, ~ \mathrm{LHS} = 3 ~\mathrm{and}~ \mathrm{RHS}=3$

$\therefore$, P(1) is true.

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

Want to prove that P(k+1) is also true, i.e., $\sum_{r=1}^{k+1} r(2r^{2}+1) = \frac{k+1}{2} (k+2)[(k+1)^{2}+k+2)] = \frac{k+1}{2} (k+2)(k^{2}+3k+3)$

$\mathrm{LHS} = r(2r^{2}+1)$

$= r(2r^{2}+1) + (k+1)[2(k+1)^{2}+1]$

$= \frac{k}{2} (k+1)(k^{2}+k+1) + (k+1)(2k^{2}+4k +3)$

$= \frac{k+1}{2} [k(k^{2}+k+1) + 2(2k^{2}+4k +3)]$

$= \frac{k+1}{2} [k^{3} + 5k^{2}+9k +6]$

$= \frac{k+1}{2} (k+2)(k^{2}+3k+3) = \mathrm{RHS}$

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

(ii)
$\mathrm{LHS}= f(r) - f(r-1)$

$= 2r^{3}+3r^{2}+r+24 - 2(r-1)^{3}-3(r-1)^{2}-(r-1)-24$

$= 2r^{3}+3r^{2}+r-2r^{3}+6r^{2}-6r+2-3r^{2}+6r-3-r+1$

$= 6r^{2}$

$\therefore, a=6$

$\sum_{r=1}^{n} r^{2}$

$= \frac{1}{6} \sum_{r=1}^{n} 6r^{2}$

$= \frac{1}{6} \sum_{r=1}^{n} f(r) - f(r-1)$

$= \frac{1}{6} [ ~f(1) - f(0)$
$~~~~~~~~~+f(2) - f(1)$
$~~~~~~~~~+f(3) - f(2)$

$~~~~~~~~~+f(n-2) - f(n-3)$
$~~~~~~~~~+f(n-1) - f(n-2)$
$~~~~~~~~~+f(n) - f(n-1)$

$= \frac{1}{6} [f(n)-f(0)]$

$= \frac{1}{6} [2n^{3}+3n^{2}+n+24-24]$

$= \frac{1}{6} [2n^{3}+3n^{2}+n]$

$= \frac{n}{6} [2n^{2}+3n+1]$

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

(iii)
$\sum_{r=1}^{n} f(r)$

$= \sum_{r=1}^{n} 2r^{3}+3r^{2}+r+24$

$= \sum_{r=1}^{n} r(2r^{2}+1) + 3 \sum_{r=1}^{n} r^{2} + \sum_{r=1}^{n} 24$

$= \frac{n}{2}(n+1)(n^{2}+n+1) + \frac{n}{2}(2n+1)(n+1) +24n$