These are the first four diagrams of a sequence.
The diagrams are made from white dots and black dots.
(a) Complete the table for Diagram 5 and Diagram 6.
(b) Write an expression, in terms of n, for the number of white dots in Diagram n.
(c) The expression for the total number of dots in Diagram n is \(\frac{1}{2}(3n^{2}-n).\)
(i) Find the total number of dots in Diagram 8.
(ii) Find an expression for the number of black dots in Diagram n.
Give your answer in its simplest form.
(d) T is the total number of dots used to make all of the first n diagrams.
\(T=an^{3}+bn^{2}\)
Find the value of a and the value of b.
You must show all your working.
▶️ Answer/Explanation
(a) Ans: 25, 36, 10, 15, 35, 51
For Diagram 5: White dots = \(5^2 = 25\), Black dots = \(\frac{5 \times 4}{2} = 10\), Total = 35.
For Diagram 6: White dots = \(6^2 = 36\), Black dots = \(\frac{6 \times 5}{2} = 15\), Total = 51.
(b) Ans: \(n^{2}\)
The number of white dots follows the pattern \(1^2, 2^2, 3^2, \dots\), so for Diagram n, it is \(n^2\).
(c)(i) Ans: 92
Substitute \(n = 8\) into \(\frac{1}{2}(3 \times 8^2 – 8) = \frac{1}{2}(192 – 8) = 92\).
(c)(ii) Ans: \(\frac{1}{2}(n^{2}-n)\)
Black dots = Total dots – White dots = \(\frac{1}{2}(3n^2 – n) – n^2 = \frac{1}{2}(n^2 – n)\).
(d) Ans: \(a = \frac{1}{2}, b = \frac{1}{2}\)
Sum of total dots for first \(n\) diagrams: \(T = \sum_{k=1}^n \frac{1}{2}(3k^2 – k)\).
Simplify to \(T = \frac{3}{2} \sum k^2 – \frac{1}{2} \sum k = \frac{3}{2} \cdot \frac{n(n+1)(2n+1)}{6} – \frac{1}{2} \cdot \frac{n(n+1)}{2}\).
After simplification, \(T = \frac{1}{2}n^3 + \frac{1}{2}n^2\). Thus, \(a = \frac{1}{2}, b = \frac{1}{2}\).
Marco is making patterns with grey and white circular mats.
The patterns form a sequence.
Marco makes a table to show some information about the patterns.
(a) Complete the table for Pattern 5.
(b) Find an expression, in terms of n, for the number of grey mats in Pattern n.
(c) Marco makes a pattern with 24 grey mats. Find the total number of mats in this pattern.
(d) Marco needs a total of 6 mats to make the first pattern. He needs a total of 16 mats to make the first two patterns. He needs a total of \(\frac{1}{6}n^{3}+an^{2}+bn\) mats to make the first n patterns. Find the value of a and the value of b.
▶️ Answer/Explanation
(a) Ans: 18, 28
For Pattern 5, grey mats follow the sequence \(3n + 3\) → \(3(5) + 3 = 18\). Total mats follow \(n^2 + 4n\) → \(5^2 + 4(5) = 25 + 20 = 45\). White mats = Total – Grey = \(45 – 18 = 27\).
(b) Ans: \(3n + 3\)
Observing the pattern: 6, 9, 12, 15, … grey mats increase by 3 each time. Thus, the expression is linear: \(3n + 3\).
(c) Ans: 45
Set \(3n + 3 = 24\) → \(n = 7\). Total mats for Pattern 7: \(7^2 + 4(7) = 49 + 28 = 77\). White mats = \(77 – 24 = 53\).
(d) Ans: \(a = \frac{3}{2}\), \(b = \frac{13}{3}\)
Using given totals: For \(n = 1\), \(\frac{1}{6} + a + b = 6\). For \(n = 2\), \(\frac{8}{6} + 4a + 2b = 16\). Solve the system to find \(a\) and \(b\).