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 $n \in \mathbb{Z}^{+}$