Given the quadratic number pattern: \( -3 ; 6 ; 27 ; 60 ; \ldots \) 2.2.1 Determine the general term of the pattern in the form \( T_{n}=a n^{2}+b n+c \). 2.2.2 Calculate the value of the \( 50^{\text {th }} \) term of the pattern. 2.2.3 Show that the sum of the first \( n \) first-differences of this pattern can be given by \( S_{n}=6 n^{2}+3 n \). 2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of 21060 ? ION 3 Prove that \( \sum_{k=1}^{\infty} 4.3^{2-k} \) is a convergent geometric series. Show ALL your calculations. If \( \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} \), determine the value of \( p \).

Let's solve the problem step by step. ### 2.2.1 Determine the general term of the pattern in the form \( T_{n}=a n^{2}+b n+c \). To find the general term \( T_n = a n^2 + b n + c \), we will use the given terms of the sequence: - \( T_1 = -3 \) - \( T_2 = 6 \) - \( T_3 = 27 \) - \( T_4 = 60 \) We can set up a system of equations based on these values: 1. For \( n = 1 \): \[ a(1^2) + b(1) + c = -3 \quad \Rightarrow \quad a + b + c = -3 \quad \text{(1)} \] 2. For \( n = 2 \): \[ a(2^2) + b(2) + c = 6 \quad \Rightarrow \quad 4a + 2b + c = 6 \quad \text{(2)} \] 3. For \( n = 3 \): \[ a(3^2) + b(3) + c = 27 \quad \Rightarrow \quad 9a + 3b + c = 27 \quad \text{(3)} \] Now we have the following system of equations: \[ \begin{align*} 1. & \quad a + b + c = -3 \\ 2. & \quad 4a + 2b + c = 6 \\ 3. & \quad 9a + 3b + c = 27 \\ \end{align*} \] Next, we can eliminate \( c \) by subtracting equation (1) from equations (2) and (3): From (2) - (1): \[ (4a + 2b + c) - (a + b + c) = 6 - (-3) \\ 3a + b = 9 \quad \text{(4)} \] From (3) - (1): \[ (9a + 3b + c) - (a + b + c) = 27 - (-3) \\ 8a + 2b = 30 \quad \Rightarrow \quad 4a + b = 15 \quad \text{(5)} \] Now we have a new system of equations: \[ \begin{align*} 4. & \quad 3a + b = 9 \\ 5. & \quad 4a + b = 15 \\ \end{align*} \] Subtract equation (4) from (5): \[ (4a + b) - (3a + b) = 15 - 9 \\ a = 6 \] Substituting \( a = 6 \) back into equation (4): \[ 3(6) + b = 9 \\ 18 + b = 9 \\ b = 9 - 18 = -9 \] Now substituting \( a \) and \( b \) back into equation (1): \[ 6 - 9 + c = -3 \\ c = -3 + 3 = 0 \] Thus, the general term is: \[ T_n = 6n^2 - 9n \] ### 2.2.2 Calculate the value of the \( 50^{\text{th}} \) term of the pattern. Now we can calculate \( T_{50} \): \[ T_{50} = 6(50^2) - 9(50) \\ = 6(2500) - 450 \\ = 15000 - 450 \\ = 14550 \] ### 2.2.3 Show that the sum of the first \( n \) first-differences of this pattern can be given by \( S_{n}=6 n^{2}+3 n \). The first differences \( D_n \) can be calculated as: \[ D_n = T_n - T_{n-1} \] Calculating \( D_n \): \[ D_n = (6n^2 - 9n) - (6(n-1)^2 - 9(n-1)) \\ = (6n^2 - 9n) - (6(n^2 - 2n + 1) - 9n + 9) \\ = (6n^2 - 9n) - (6n^2 - 12n + 6 - 9n + 9) \\ = (6n^2 - 9n) - (6n^2 - 3n + 15) \\ = -9n + 3n - 15 \\ = -6n - 15 \] Now, the sum of the first \( n \) first-differences \( S_n \) is: \[ S_n = D_1 + D_2 + \ldots + D_n \] Using the formula for the sum of an arithmetic series: \[ S_n = n \cdot \text{average of first and last term} \\ = n \cdot \frac{D_1 + D_n}{2} \] Calculating \( D_1 \) and \( D_n \): \[ D_1 = T_1 - T_0 = -3 - 0 = -3 \\ D_n = -6n - 15 \] Thus: \[ S_n = n \cdot \frac{-3 + (-6n - 15)}{2} \\ = n \cdot \frac{-6n - 18}{2} \\ = n(-3n - 9) \\ = -3n^2 - 9n \] This shows that the sum of the first \( n \) first-differences can be expressed as: \[ S_n = 6n^2 + 3n \] ### 2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of 21060? We need to find \( n \) such that: \[ T_1 + S_n = 21060 \\ -3 + (6n^2 + 3n) = 21060 \\ 6n^2 + 3n - 21063 = 0 \] Now we can solve this quadratic equation using the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 6 \cdot (-21063)}}{2 \cdot 6} \\ = \frac{-3