7. Find an expression for the $$ n $$th term of each of the quadratic sequences: $$ \begin{array}{lll} ext { (i) } 7,10,15,22,31, \ldots & ext { (ii) } 2,7,14,23, \ldots & ext { (iii) } 2,8,18,32,50, \ldots\end{array} $$ 8. Show that the $$ n $$th term of the quadratic sequence $$ 8,15,26,41,60, \ldots $$ is given by $$ T_{n}=2 n^{2}+n+5 $$

Question

7. Find an expression for the $$ n $$th term of each of the quadratic sequences: $$ \begin{array}{lll}	ext { (i) } 7,10,15,22,31, \ldots & 	ext { (ii) } 2,7,14,23, \ldots & 	ext { (iii) } 2,8,18,32,50, \ldots\end{array} $$ 8. Show that the $$ n $$th term of the quadratic sequence $$ 8,15,26,41,60, \ldots $$ is given by $$ T_{n}=2 n^{2}+n+5 $$

UpStudy AI · Accepted Answer

Sure, let's tackle both problems step by step.

---

## **Problem 7**

**Find an expression for the $$ n $$th term of each of the quadratic sequences:**

### **(i) Sequence:** $$ 7, 10, 15, 22, 31, \ldots $$

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  10 - 7 = 3,\quad 15 - 10 = 5,\quad 22 - 15 = 7,\quad 31 - 22 = 9,\ldots
  $$
  Sequence: $$ 3, 5, 7, 9, \ldots $$

- **Second Differences:**
  $$
  5 - 3 = 2,\quad 7 - 5 = 2,\quad 9 - 7 = 2,\ldots
  $$
  The second differences are constant ($$ 2 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

For a quadratic sequence, the general term is:
$$
T_n = an^2 + bn + c
$$
Given that the second difference is $$ 2 $$, we know:
$$
2a = 2 \quad \Rightarrow \quad a = 1
$$
So, the general form simplifies to:
$$
T_n = n^2 + bn + c
$$
Using the first term ($$ n = 1 $$):
$$
7 = 1^2 + b(1) + c \quad \Rightarrow \quad b + c = 6 \quad \text{(Equation 1)}
$$
Using the second term ($$ n = 2 $$):
$$
10 = 2^2 + b(2) + c \quad \Rightarrow \quad 4 + 2b + c = 10 \quad \Rightarrow \quad 2b + c = 6 \quad \text{(Equation 2)}
$$
Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 6 - 6 \quad \Rightarrow \quad b = 0
$$
Substitute $$ b = 0 $$ into Equation 1:
$$
0 + c = 6 \quad \Rightarrow \quad c = 6
$$
**Final Formula:**
$$
T_n = n^2 + 6
$$

---

### **(ii) Sequence:** $$ 2, 7, 14, 23, \ldots $$

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  7 - 2 = 5,\quad 14 - 7 = 7,\quad 23 - 14 = 9,\ldots
  $$
  Sequence: $$ 5, 7, 9, \ldots $$

- **Second Differences:**
  $$
  7 - 5 = 2,\quad 9 - 7 = 2,\ldots
  $$
  The second differences are constant ($$ 2 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

$$
T_n = an^2 + bn + c
$$
With $$ 2a = 2 $$, we have $$ a = 1 $$:
$$
T_n = n^2 + bn + c
$$
Using the first term ($$ n = 1 $$):
$$
2 = 1^2 + b(1) + c \quad \Rightarrow \quad b + c = 1 \quad \text{(Equation 1)}
$$
Using the second term ($$ n = 2 $$):
$$
7 = 2^2 + b(2) + c \quad \Rightarrow \quad 4 + 2b + c = 7 \quad \Rightarrow \quad 2b + c = 3 \quad \text{(Equation 2)}
$$
Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 3 - 1 \quad \Rightarrow \quad b = 2
$$
Substitute $$ b = 2 $$ into Equation 1:
$$
2 + c = 1 \quad \Rightarrow \quad c = -1
$$
**Final Formula:**
$$
T_n = n^2 + 2n - 1
$$

---

### **(iii) Sequence:** $$ 2, 8, 18, 32, 50, \ldots $$

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  8 - 2 = 6,\quad 18 - 8 = 10,\quad 32 - 18 = 14,\quad 50 - 32 = 18,\ldots
  $$
  Sequence: $$ 6, 10, 14, 18, \ldots $$

- **Second Differences:**
  $$
  10 - 6 = 4,\quad 14 - 10 = 4,\quad 18 - 14 = 4,\ldots
  $$
  The second differences are constant ($$ 4 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

$$
T_n = an^2 + bn + c
$$
With $$ 2a = 4 $$, we have $$ a = 2 $$:
$$
T_n = 2n^2 + bn + c
$$
Using the first term ($$ n = 1 $$):
$$
2 = 2(1)^2 + b(1) + c \quad \Rightarrow \quad 2 + b + c = 2 \quad \Rightarrow \quad b + c = 0 \quad \text{(Equation 1)}
$$
Using the second term ($$ n = 2 $$):
$$
8 = 2(2)^2 + b(2) + c \quad \Rightarrow \quad 8 + 2b + c = 8 \quad \Rightarrow \quad 2b + c = 0 \quad \text{(Equation 2)}
$$
Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 0 - 0 \quad \Rightarrow \quad b = 0
$$
Substitute $$ b = 0 $$ into Equation 1:
$$
0 + c = 0 \quad \Rightarrow \quad c = 0
$$
**Final Formula:**
$$
T_n = 2n^2
$$

---

## **Problem 8**

**Show that the $$ n $$th term of the quadratic sequence $$ 8, 15, 26, 41, 60, \ldots $$ is given by $$ T_{n}=2 n^{2}+n+5 $$.**

**Step 1: Determine the Differences**

- **First Differences:**
  $$
  15 - 8 = 7,\quad 26 - 15 = 11,\quad 41 - 26 = 15,\quad 60 - 41 = 19,\ldots
  $$
  Sequence: $$ 7, 11, 15, 19, \ldots $$

- **Second Differences:**
  $$
  11 - 7 = 4,\quad 15 - 11 = 4,\quad 19 - 15 = 4,\ldots
  $$
  The second differences are constant ($$ 4 $$), indicating a quadratic sequence.

**Step 2: Formulate the General Term**

Assume the general term is:
$$
T_n = an^2 + bn + c
$$
Given that the second difference is $$ 4 $$, we have:
$$
2a = 4 \quad \Rightarrow \quad a = 2
$$
So, the general form becomes:
$$
T_n = 2n^2 + bn + c
$$

**Step 3: Use Given Terms to Find $$ b $$ and $$ c $$**

Using the first three terms:

1. For $$ n = 1 $$:
   $$
   8 = 2(1)^2 + b(1) + c \quad \Rightarrow \quad 2 + b + c = 8 \quad \Rightarrow \quad b + c = 6 \quad \text{(Equation 1)}
   $$

2. For $$ n = 2 $$:
   $$
   15 = 2(2)^2 + b(2) + c \quad \Rightarrow \quad 8 + 2b + c = 15 \quad \Rightarrow \quad 2b + c = 7 \quad \text{(Equation 2)}
   $$

3. For $$ n = 3 $$:
   $$
   26 = 2(3)^2 + b(3) + c \quad \Rightarrow \quad 18 + 3b + c = 26 \quad \Rightarrow \quad 3b + c = 8 \quad \text{(Equation 3)}
   $$

**Step 4: Solve the System of Equations**

Subtract Equation 1 from Equation 2:
$$
2b + c - (b + c) = 7 - 6 \quad \Rightarrow \quad b = 1
$$
Substitute $$ b = 1 $$ into Equation 1:
$$
1 + c = 6 \quad \Rightarrow \quad c = 5
$$

**Final Formula:**
$$
T_n = 2n^2 + n + 5
$$

**Verification:**
$$
\begin{align*}
n = 1: & \quad 2(1)^2 + 1 + 5 = 2 + 1 + 5 = 8 \
n = 2: & \quad 2(2)^2 + 2 + 5 = 8 + 2 + 5 = 15 \
n = 3: & \quad 2(3)^2 + 3 + 5 = 18 + 3 + 5 = 26 \
n = 4: & \quad 2(4)^2 + 4 + 5 = 32 + 4 + 5 = 41 \
n = 5: & \quad 2(5)^2 + 5 + 5 = 50 + 5 + 5 = 60 \
\end{align*}
$$
All terms match the given sequence.

---

**Summary of Answers:**

- **Problem 7:**
  - **(i)** $$ T_n = n^2 + 6 $$
  - **(ii)** $$ T_n = n^2 + 2n - 1 $$
  - **(iii)** $$ T_n = 2n^2 $$

- **Problem 8:**
  
  The $$ n $$th term of the sequence $$ 8,15,26,41,60,\ldots $$ is $$ T_n = 2n^2 + n + 5 $$.
**Problem 7 Solutions:** - **(i)** $$ T_n = n^2 + 6 $$ - **(ii)** $$ T_n = n^2 + 2n - 1 $$ - **(iii)** $$ T_n = 2n^2 $$ **Problem 8 Solution:** The $$ n $$th term is $$ T_n = 2n^2 + n + 5 $$.

Solución de tutoría real

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Pre Algebra Questions