The number pattern $$ 7 ; 22 ; 47 ; 82 ; 127 ; \ldots \ldots $$. is given such that it continues consistently. Determine: 4.1 The first difference 4.2 The second difference, what do you notice? 4.3 The formula for the $$ n $$-th term.

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Let's analyze the given number pattern: $$ 7, 22, 47, 82, 127, \ldots $$

### 4.1 The First Difference

The **first difference** is the difference between consecutive terms in the sequence.

$$
\begin{align*}
22 - 7 &= 15 \
47 - 22 &= 25 \
82 - 47 &= 35 \
127 - 82 &= 45 \
\end{align*}
$$

So, the first differences are:
$$
15, \, 25, \, 35, \, 45, \, \ldots
$$

### 4.2 The Second Difference

The **second difference** is the difference between consecutive first differences.

$$
\begin{align*}
25 - 15 &= 10 \
35 - 25 &= 10 \
45 - 35 &= 10 \
\end{align*}
$$

So, the second differences are:
$$
10, \, 10, \, 10, \ldots
$$

**Observation:** The second differences are constant ($$10$$). This indicates that the original sequence is **quadratic**.

### 4.3 The Formula for the $$ n $$-th Term

Since the sequence is quadratic, the $$ n $$-th term can be expressed in the form:
$$
a n^2 + b n + c
$$

To find the coefficients $$ a $$, $$ b $$, and $$ c $$, we can set up the following equations using the first three terms of the sequence:

1. For $$ n = 1 $$:
$$
a(1)^2 + b(1) + c = 7 \quad \Rightarrow \quad a + b + c = 7
$$
2. For $$ n = 2 $$:
$$
a(2)^2 + b(2) + c = 22 \quad \Rightarrow \quad 4a + 2b + c = 22
$$
3. For $$ n = 3 $$:
$$
a(3)^2 + b(3) + c = 47 \quad \Rightarrow \quad 9a + 3b + c = 47
$$

**Solving the System of Equations:**

Subtract the first equation from the second:
$$
(4a + 2b + c) - (a + b + c) = 22 - 7 \quad \Rightarrow \quad 3a + b = 15 \quad \text{(Equation 4)}
$$

Subtract the second equation from the third:
$$
(9a + 3b + c) - (4a + 2b + c) = 47 - 22 \quad \Rightarrow \quad 5a + b = 25 \quad \text{(Equation 5)}
$$

Subtract Equation 4 from Equation 5:
$$
(5a + b) - (3a + b) = 25 - 15 \quad \Rightarrow \quad 2a = 10 \quad \Rightarrow \quad a = 5
$$

Substitute $$ a = 5 $$ into Equation 4:
$$
3(5) + b = 15 \quad \Rightarrow \quad 15 + b = 15 \quad \Rightarrow \quad b = 0
$$

Substitute $$ a = 5 $$ and $$ b = 0 $$ into the first equation:
$$
5 + 0 + c = 7 \quad \Rightarrow \quad c = 2
$$

**Therefore, the formula for the $$ n $$-th term is:**
$$
\boxed{a_n = 5n^2 + 2}
$$

**Verification:**
$$
\begin{align*}
n = 1 &: \quad 5(1)^2 + 2 = 7 \
n = 2 &: \quad 5(2)^2 + 2 = 22 \
n = 3 &: \quad 5(3)^2 + 2 = 47 \
n = 4 &: \quad 5(4)^2 + 2 = 82 \
n = 5 &: \quad 5(5)^2 + 2 = 127 \
\end{align*}
$$
All values match the given sequence.
The first difference is 15, the second difference is 10, and the $$ n $$-th term formula is $$ 5n^2 + 2 $$.

The number pattern \( 7 ; 22 ; 47 ; 82 ; 127 ; \ldots \ldots \). is given such that it continues consistently. Determine: 4.1 The first difference 4.2 The second difference, what do you notice? 4.3 The formula for the \( n \)-th term.

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