(a) Triangular Numbers:

(i) Verify \( P_3(n) = \frac{n(n+1)}{2} \):

Substitute \( r = 3 \):

\[ P_3(n) = \frac{(3-2)n^2 – (3-4)n}{2} = \frac{n^2 + n}{2} \quad (M1) \]

\[ = \frac{n(n+1)}{2} \quad (A1) \]

[2 marks]

(ii) Find \( n \) for \( P_3(n) = 351 \):

\[ \frac{n(n+1)}{2} = 351 \implies n^2 + n – 702 = 0 \quad (M1) \]

Solve: \( n = \frac{-1 \pm \sqrt{1 + 2808}}{2} = \frac{-1 \pm 53}{2} \), so \( n = 26 \) (discard negative) (A1).

351 is the 26th triangular number.

[2 marks]

(b) Sum of Consecutive Triangular Numbers:

(i) Show \( P_3(n) + P_3(n+1) = (n+1)^2 \):

\[ P_3(n) = \frac{n(n+1)}{2}, \quad P_3(n+1) = \frac{(n+1)(n+2)}{2} \quad (M1) \]

\[ P_3(n) + P_3(n+1) = \frac{n(n+1) + (n+1)(n+2)}{2} = \frac{(n+1)(n + n + 2)}{2} = \frac{(n+1)(2n+2)}{2} \quad (A1) \]

\[ = (n+1)^2 \quad (A1) \]

[3 marks]

(ii) The sum of the \( n \)-th and \( (n+1) \)-th triangular numbers is the \( (n+1) \)-th square number (A1).

[1 mark]

(iii) For \( n = 4 \), diagram showing \( P_3(4) + P_3(5) = 5^2 \):

\[ P_3(4) = 10, \quad P_3(5) = 15, \quad 10 + 15 = 25 = 5^2 \]

Sketch a 5×5 grid with 10 dots (triangle) and 15 dots (triangle) forming a 25-dot square (M1, A1).

[2 marks]

(c) Show \( 8P_3(n) + 1 \) is an Odd Square:

\[ 8P_3(n) + 1 = 8 \cdot \frac{n(n+1)}{2} + 1 = 4n(n+1) + 1 \quad (M1) \]

\[ = 4n^2 + 4n + 1 = (2n + 1)^2 \quad (A1) \]

\( 2n + 1 \) is odd for \( n \in \mathbb{Z}^+ \) (A1).

[3 marks]

(d) Pentagonal Number Formula:

\[ P_5(n) = 1 + 4 + 7 + \dots + (3n – 2) \], arithmetic series with \( u_1 = 1 \), \( d = 3 \), \( n \) terms (M1).

\[ P_5(n) = \frac{n}{2} [2 \cdot 1 + (n-1) \cdot 3] = \frac{n}{2} (2 + 3n – 3) \quad (A1) \]

\[ = \frac{n(3n – 1)}{2} \quad (A1) \]

[3 marks]

(e) Smallest Number Both Triangular and Pentagonal:

Equate: \( \frac{n(n+1)}{2} = \frac{m(3m-1)}{2} \implies n(n+1) = m(3m-1) \quad (M1) \).

Let \( m = n – k \), so:

\[ n(n+1) = (n-k)(3(n-k)-1) \implies 2n^2 – 2(3k+1)n + (3k^2 + k) = 0 \quad (A1) \]

Discriminant: \( \Delta = 4(3k+1)^2 – 8(3k^2 + k) = 4(k+1)(3k+1) \), must be a perfect square (A1).

For \( k = 8 \), solve: \( 2n^2 – 50n + 200 = 0 \implies n = 5, 20 \). Smallest \( n > 1 \): 20, gives 210 (A1).

[4 marks]

(f) Prove by Induction:

Base case (\( n = 1 \)):

\[ P_r(1) = 1, \quad \frac{(r-2) \cdot 1^2 – (r-4) \cdot 1}{2} = \frac{r-2 – r + 4}{2} = 1 \quad (M1) \]

Assume true for \( n = k \): \( P_r(k) = \frac{(r-2)k^2 – (r-4)k}{2} \quad (A1) \).

For \( n = k+1 \):

\[ P_r(k+1) = P_r(k) + [1 + k(r-2)] \quad (M1) \]

\[ = \frac{(r-2)k^2 – (r-4)k + 2 + 2k(r-2)}{2} = \frac{(r-2)(k^2 + 2k + 1) – (r-4)k + 2}{2} \quad (A1) \]

\[ = \frac{(r-2)(k+1)^2 – (r-4)(k+1)}{2} \quad (A1) \]

True for all \( n \in \mathbb{Z}^+ \).

[4 marks]