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

Question

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

UpStudy AI · Accepted Answer

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:**
   $$
   a_n = -n^2 - \frac{1}{2}n + \frac{1}{2}
   $$
The general terms for the quadratic number patterns are: 1. $$ a_n = n^2 + 2n + 5 $$ 2. $$ a_n = -3n^2 + 2n - 1 $$ 3. $$ a_n = \frac{1}{2}n^2 + n - \frac{3}{2} $$ 4. $$ a_n = -n^2 - \frac{1}{2}n + \frac{1}{2} $$

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