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

KS Comments:

Students mostly did not struggle with this question. The instructions were clear. Some students found it difficult to relate (iii) to (i) and were unsure of how to resolve the summation

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