1. **Stating the problem:** We have a sequence of patterns made with discs. We want to find formulas for the number of discs in pattern $n$, the total discs for the first $n$ patterns, and solve related questions. 2. **Given data:** The number of discs for patterns 1 to 6 are: 1, 4, 4, 10, 13, 16. We see the sequence is not strictly arithmetic but the user gave a formula $T_n = 3n - 2$ for the number of discs in pattern $n$. 3. **Formula for $T_n$ (number of discs in pattern $n$):** $$ T_n = 3n - 2 $$ This is an arithmetic sequence with first term $a=1$ and common difference $d=3$. 4. **(e)(i) Find formula for $S_n$, total discs for first $n$ patterns:** The sum of the first $n$ terms of an arithmetic sequence is: $$ S_n = \frac{n}{2} (2a + (n-1)d) $$ Substitute $a=1$, $d=3$: $$ S_n = \frac{n}{2} (2(1) + (n-1)3) = \frac{n}{2} (2 + 3n - 3) = \frac{n}{2} (3n - 1) $$ So, $$ \boxed{S_n = \frac{n(3n - 1)}{2}} $$ 5. **(e)(ii) Find total number of complete patterns with 477 discs:** We want the largest integer $n$ such that $$ S_n \leq 477 $$ Substitute $S_n$: $$ \frac{n(3n - 1)}{2} \leq 477 $$ Multiply both sides by 2: $$ n(3n - 1) \leq 954 $$ Expand: $$ 3n^2 - n - 954 \leq 0 $$ Solve quadratic inequality: $$ 3n^2 - n - 954 = 0 $$ Use quadratic formula: $$ n = \frac{1 \pm \sqrt{(-1)^2 - 4(3)(-954)}}{2 \times 3} = \frac{1 \pm \sqrt{1 + 11448}}{6} = \frac{1 \pm \sqrt{11449}}{6} $$ Calculate $ \sqrt{11449} \approx 107.02$ So, $$ n = \frac{1 \pm 107.02}{6} $$ Positive root: $$ n = \frac{1 + 107.02}{6} = \frac{108.02}{6} = 18.003 $$ Since $n$ must be an integer number of patterns, the maximum $n$ is 18. **Final answers:** - Formula for total discs in first $n$ patterns: $$ S_n = \frac{n(3n - 1)}{2} $$ - Maximum complete patterns with 477 discs: $$ \boxed{18} $$