The following information is given about a quadratic number pattern: $$ T_{1}=3, T_{2}-T_{1}=9 $$ and $$ T_{3}-T_{2}=21 $$ $$ 2.2 .1 \quad $$ Show that $$ T_{5}=111 $$ $$ 2.2 .2 \quad $$ Show that the general term of the quadratic pattern is $$ T_{n}=6 n^{2}-9 n+6 $$ $$ 2.2 .3 \quad $$ Show that the pattern is increasing for all $$ n \in N $$.

Question

UpStudy AI · Accepted Answer

Let's analyze the given quadratic number pattern step by step to address each part of the problem.

### Given:
- $$ T_1 = 3 $$
- $$ T_2 - T_1 = 9 $$ ⇒ $$ T_2 = T_1 + 9 = 3 + 9 = 12 $$
- $$ T_3 - T_2 = 21 $$ ⇒ $$ T_3 = T_2 + 21 = 12 + 21 = 33 $$

### 2.2.1. Show that $$ T_5 = 111 $$

Since the pattern is quadratic, the second differences between consecutive terms are constant. Let's determine the second differences and use them to find $$ T_4 $$ and $$ T_5 $$.

1. **First Differences:**
   - $$ T_2 - T_1 = 12 - 3 = 9 $$
   - $$ T_3 - T_2 = 33 - 12 = 21 $$

2. **Second Difference:**
   - $$ \Delta_2 = (T_3 - T_2) - (T_2 - T_1) = 21 - 9 = 12 $$

3. **Continuing the Pattern:**
   - **Difference $$ T_4 - T_3 $$:**
     $$
     \Delta_3 = \Delta_2 + 12 = 21 + 12 = 33
     $$
     $$
     T_4 = T_3 + \Delta_3 = 33 + 33 = 66
     $$
   
   - **Difference $$ T_5 - T_4 $$:**
     $$
     \Delta_4 = \Delta_3 + 12 = 33 + 12 = 45
     $$
     $$
     T_5 = T_4 + \Delta_4 = 66 + 45 = 111
     $$

**Therefore, $$ T_5 = 111 $$.**

---

### 2.2.2. Show that the general term of the quadratic pattern is $$ T_n = 6n^2 - 9n + 6 $$

To find the general term $$ T_n $$ of a quadratic sequence, we assume it has the form:
$$
T_n = an^2 + bn + c
$$
We can set up equations using the known terms:

1. **Using $$ n = 1 $$:**
   $$
   6(1)^2 - 9(1) + 6 = 6 - 9 + 6 = 3 = T_1
   $$

2. **Using $$ n = 2 $$:**
   $$
   6(2)^2 - 9(2) + 6 = 24 - 18 + 6 = 12 = T_2
   $$

3. **Using $$ n = 3 $$:**
   $$
   6(3)^2 - 9(3) + 6 = 54 - 27 + 6 = 33 = T_3
   $$

Since the general form $$ T_n = 6n^2 - 9n + 6 $$ correctly produces the given terms, we can confirm that this is indeed the general term of the quadratic pattern.

**Therefore, the general term is $$ T_n = 6n^2 - 9n + 6 $$.**

---

### 2.2.3. Show that the pattern is increasing for all $$ n \in \mathbb{N} $$

To determine if the sequence is increasing for all natural numbers $$ n $$, we need to show that $$ T_{n+1} > T_n $$ for all $$ n $$.

1. **Compute $$ T_{n+1} - T_n $$:**
   $$
   T_{n+1} = 6(n+1)^2 - 9(n+1) + 6 = 6n^2 + 12n + 6 - 9n - 9 + 6 = 6n^2 + 3n + 3
   $$
   $$
   T_n = 6n^2 - 9n + 6
   $$
   $$
   T_{n+1} - T_n = (6n^2 + 3n + 3) - (6n^2 - 9n + 6) = 12n - 3
   $$

2. **Analyze $$ T_{n+1} - T_n $$:**
   $$
   T_{n+1} - T_n = 12n - 3
   $$
   For $$ n \in \mathbb{N} $$ (i.e., $$ n \geq 1 $$):
   $$
   12n - 3 \geq 12(1) - 3 = 9 > 0
   $$
   As $$ n $$ increases, $$ 12n - 3 $$ becomes larger, ensuring that $$ T_{n+1} - T_n > 0 $$.

**Therefore, the sequence is increasing for all $$ n \in \mathbb{N} $$.**
$$ T_5 = 111 $$, the general term is $$ T_n = 6n^2 - 9n + 6 $$, and the pattern is increasing for all natural numbers $$ n $$.

The following information is given about a quadratic number pattern:
and
Show that
Show that the general term of the quadratic pattern is
Show that the pattern is increasing for all .

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Algebra Questions

The following information is given about a quadratic number pattern: and Show that Show that the general term of the quadratic pattern is Show that the pattern is increasing for all .