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

(i)
u_2 = u_{1+1} = \frac{3u_1 - 1}{6} = \frac{5}{6}

u_3 = \frac{3u_2 -1}{6} = \frac{1}{4}

(ii)
As n \rightarrow \infty, u_n \rightarrow l, u_{n+1} \rightarrow l

\Rightarrow l = \frac{3l-1}{6}

6l = 3l - 1

l = -\frac{1}{3}

(iii)
Let P(n) be the statement “u_n = \frac{14}{3} (0.5)^{n} - \frac{1}{3}” for n \in {\mathbb{Z}}^{+}

When n = 1, \mathrm{LHS} = u_1 = 2,~ \mathrm{RHS} = \frac{14}{3}(0.5) - \frac{1}{3} = 2 = \mathrm{LHS}.

Thus, P(1) is true.

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

u_k = \frac{14}{3} (0.5)^{k} - \frac{1}{3}

To prove P(k+1) is true, i.e.

u_{k+1} = \frac{14}{3} (0.5)^{k+1} - \frac{1}{3}

\mathrm{LHS}

= u_{k+1}

= \frac{3u_{k}-1}{6}

= \frac{1}{6} [14 (0.5)^{k} - 1-1]

= \frac{1}{6} [14 (0.5)^{k} - 1-1]

= \frac{14}{6} 2 (0.5)^{k+1} - \frac{1}{3}

= \frac{14}{3} (0.5)^{k+1} - \frac{1}{3}

= \mathrm{RHS}

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

KS Comments:

This question is quite straight forward. Some students did not realise to use the given recurrence relation to solve the MI.

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