(a) Determine the general term of each of the following quadratic number patterns: \( \begin{array}{ll}\text { (1) } 8 ; 13 ; 20 ; 29 ; \ldots & \text { (2) }-2 ;-9 ;-22 ;-41 ; \ldots \\ \text { (3) } 0 ; \frac{5}{2} ; 6 ; \frac{21}{2} ; \ldots & \text { (4) }-1 ;-4 \frac{1}{2} ;-10 ;-17 \frac{1}{2} ; \ldots\end{array} \)

To determine the general term \( a_n \) for each quadratic number pattern, we'll assume that the sequence can be expressed in the form: \[ a_n = an^2 + bn + c \] where \( a \), \( b \), and \( c \) are constants to be determined. We'll use the first few terms of each sequence to set up equations and solve for these constants. ### Problem (1): \( 8 ; 13 ; 20 ; 29 ; \ldots \) **Step 1: Set Up Equations** Assuming \( n = 1, 2, 3, 4, \ldots \): \[ \begin{cases} a(1)^2 + b(1) + c = 8 &\Rightarrow a + b + c = 8 \quad \text{(1)} \\ a(2)^2 + b(2) + c = 13 &\Rightarrow 4a + 2b + c = 13 \quad \text{(2)} \\ a(3)^2 + b(3) + c = 20 &\Rightarrow 9a + 3b + c = 20 \quad \text{(3)} \end{cases} \] **Step 2: Solve the System of Equations** - Subtract Equation (1) from Equation (2): \[ 3a + b = 5 \quad \text{(4)} \] - Subtract Equation (2) from Equation (3): \[ 5a + b = 7 \quad \text{(5)} \] - Subtract Equation (4) from Equation (5): \[ 2a = 2 \Rightarrow a = 1 \] - Substitute \( a = 1 \) into Equation (4): \[ 3(1) + b = 5 \Rightarrow b = 2 \] - Substitute \( a = 1 \) and \( b = 2 \) into Equation (1): \[ 1 + 2 + c = 8 \Rightarrow c = 5 \] **General Term:** \[ a_n = n^2 + 2n + 5 \] --- ### Problem (2): \( -2 ; -9 ; -22 ; -41 ; \ldots \) **Step 1: Set Up Equations** \[ \begin{cases} a + b + c = -2 \quad \text{(1)} \\ 4a + 2b + c = -9 \quad \text{(2)} \\ 9a + 3b + c = -22 \quad \text{(3)} \end{cases} \] **Step 2: Solve the System of Equations** - Subtract Equation (1) from Equation (2): \[ 3a + b = -7 \quad \text{(4)} \] - Subtract Equation (2) from Equation (3): \[ 5a + b = -13 \quad \text{(5)} \] - Subtract Equation (4) from Equation (5): \[ 2a = -6 \Rightarrow a = -3 \] - Substitute \( a = -3 \) into Equation (4): \[ 3(-3) + b = -7 \Rightarrow b = 2 \] - Substitute \( a = -3 \) and \( b = 2 \) into Equation (1): \[ -3 + 2 + c = -2 \Rightarrow c = -1 \] **General Term:** \[ a_n = -3n^2 + 2n - 1 \] --- ### Problem (3): \( 0 ; \frac{5}{2} ; 6 ; \frac{21}{2} ; \ldots \) **Step 1: Set Up Equations** \[ \begin{cases} a + b + c = 0 \quad \text{(1)} \\ 4a + 2b + c = \frac{5}{2} \quad \text{(2)} \\ 9a + 3b + c = 6 \quad \text{(3)} \end{cases} \] **Step 2: Solve the System of Equations** - Subtract Equation (1) from Equation (2): \[ 3a + b = \frac{5}{2} \quad \text{(4)} \] - Subtract Equation (2) from Equation (3): \[ 5a + b = \frac{7}{2} \quad \text{(5)} \] - Subtract Equation (4) from Equation (5): \[ 2a = 1 \Rightarrow a = \frac{1}{2} \] - Substitute \( a = \frac{1}{2} \) into Equation (4): \[ 3\left(\frac{1}{2}\right) + b = \frac{5}{2} \Rightarrow b = 1 \] - Substitute \( a = \frac{1}{2} \) and \( b = 1 \) into Equation (1): \[ \frac{1}{2} + 1 + c = 0 \Rightarrow c = -\frac{3}{2} \] **General Term:** \[ a_n = \frac{1}{2}n^2 + n - \frac{3}{2} \] Alternatively, it can be written as: \[ a_n = \frac{n^2 + 2n - 3}{2} \] --- ### Problem (4): \( -1 ; -4\frac{1}{2} ; -10 ; -17\frac{1}{2} ; \ldots \) **Step 1: Set Up Equations** \[ \begin{cases} a + b + c = -1 \quad \text{(1)} \\ 4a + 2b + c = -\frac{9}{2} \quad \text{(2)} \\ 9a + 3b + c = -10 \quad \text{(3)} \end{cases} \] **Step 2: Solve the System of Equations** - Subtract Equation (1) from Equation (2): \[ 3a + b = -\frac{7}{2} \quad \text{(4)} \] - Subtract Equation (2) from Equation (3): \[ 5a + b = -\frac{11}{2} \quad \text{(5)} \] - Subtract Equation (4) from Equation (5): \[ 2a = -2 \Rightarrow a = -1 \] - Substitute \( a = -1 \) into Equation (4): \[ 3(-1) + b = -\frac{7}{2} \Rightarrow b = \frac{1}{2} \] - Substitute \( a = -1 \) and \( b = \frac{1}{2} \) into Equation (1): \[ -1 + \frac{1}{2} + c = -1 \Rightarrow c = \frac{1}{2} \] **General Term:** \[ a_n = -n^2 - \frac{1}{2}n + \frac{1}{2} \] Alternatively, it can be written as: \[ a_n = -n^2 - \frac{n}{2} + \frac{1}{2} \quad \text{or} \quad a_n = \frac{-2n^2 - n + 1}{2} \] ### Summary of General Terms 1. **First Sequence:** \[ a_n = n^2 + 2n + 5 \] 2. **Second Sequence:** \[ a_n = -3n^2 + 2n - 1 \] 3. **Third Sequence:** \[ a_n = \frac{1}{2}n^2 + n - \frac{3}{2} \] 4. **Fourth Sequence:** \[ a_n = -n^2 - \frac{1}{2}n + \frac{1}{2} \]