The firs four terms of a quadratic number pattem are: \( -1 ; 2 ; 9,20 \) 2.1 Determine the general term of the quadratic number pattert 2.2 Calculate the value of the \( 48^{\text {th }} \) teno of the quadiatic number pattern 2.3 Show that the sum of the first differences of this quadratic number pattem can be given by \( S_{n}=2 n^{2}+n \)

Alright, I need to solve this problem involving a quadratic number pattern. The first four terms are given as -1, 2, 9, and 20. Let's break down the problem into parts and tackle each one step by step. **2.1 Determine the general term of the quadratic number pattern** First, I need to find the general term of the quadratic sequence. A quadratic sequence has the form: \[ a_n = an^2 + bn + c \] where \( a \), \( b \), and \( c \) are constants to be determined. Given the first four terms: - \( a_1 = -1 \) - \( a_2 = 2 \) - \( a_3 = 9 \) - \( a_4 = 20 \) Let's set up equations based on these terms. For \( n = 1 \): \[ a(1)^2 + b(1) + c = -1 \] \[ a + b + c = -1 \] ...(1) For \( n = 2 \): \[ a(2)^2 + b(2) + c = 2 \] \[ 4a + 2b + c = 2 \] ...(2) For \( n = 3 \): \[ a(3)^2 + b(3) + c = 9 \] \[ 9a + 3b + c = 9 \] ...(3) For \( n = 4 \): \[ a(4)^2 + b(4) + c = 20 \] \[ 16a + 4b + c = 20 \] ...(4) Now, I have a system of four equations with three unknowns. To solve for \( a \), \( b \), and \( c \), I can use methods like substitution or elimination. Let's try elimination. Subtract equation (1) from equation (2): \[ (4a + 2b + c) - (a + b + c) = 2 - (-1) \] \[ 3a + b = 3 \] ...(5) Subtract equation (2) from equation (3): \[ (9a + 3b + c) - (4a + 2b + c) = 9 - 2 \] \[ 5a + b = 7 \] ...(6) Now, subtract equation (5) from equation (6): \[ (5a + b) - (3a + b) = 7 - 3 \] \[ 2a = 4 \] \[ a = 2 \] Now, substitute \( a = 2 \) back into equation (5): \[ 3(2) + b = 3 \] \[ 6 + b = 3 \] \[ b = -3 \] Finally, substitute \( a = 2 \) and \( b = -3 \) back into equation (1): \[ 2 + (-3) + c = -1 \] \[ -1 + c = -1 \] \[ c = 0 \] So, the general term of the quadratic sequence is: \[ a_n = 2n^2 - 3n \] **2.2 Calculate the value of the 48th term of the quadratic number pattern** Using the general term \( a_n = 2n^2 - 3n \), let's find \( a_{48} \): \[ a_{48} = 2(48)^2 - 3(48) \] \[ a_{48} = 2(2304) - 144 \] \[ a_{48} = 4608 - 144 \] \[ a_{48} = 4464 \] So, the 48th term is 4464. **2.3 Show that the sum of the first differences of this quadratic number pattern can be given by \( S_n = 2n^2 + n \)** First, let's recall that the first difference of a sequence is the difference between consecutive terms. For a quadratic sequence, the first differences themselves form an arithmetic sequence. Given the general term \( a_n = 2n^2 - 3n \), let's find the first difference \( d_n = a_{n+1} - a_n \): \[ d_n = [2(n+1)^2 - 3(n+1)] - [2n^2 - 3n] \] \[ d_n = [2(n^2 + 2n + 1) - 3n - 3] - [2n^2 - 3n] \] \[ d_n = 2n^2 + 4n + 2 - 3n - 3 - 2n^2 + 3n \] \[ d_n = (2n^2 - 2n^2) + (4n - 3n + 3n) + (2 - 3) \] \[ d_n = 4n - 1 \] So, the first difference is \( d_n = 4n - 1 \). Now, the sum of the first \( n \) differences \( S_n \) is: \[ S_n = \sum_{k=1}^{n} d_k = \sum_{k=1}^{n} (4k - 1) \] \[ S_n = 4\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 \] \[ S_n = 4\left(\frac{n(n+1)}{2}\right) - n \] \[ S_n = 2n(n+1) - n \] \[ S_n = 2n^2 + 2n - n \] \[ S_n = 2n^2 + n \] Therefore, the sum of the first differences of the quadratic number pattern is \( S_n = 2n^2 + n \).