The firs four terms of a quadratic number pattem are: $$ -1 ; 2 ; 9,20 $$ 2.1 Determine the general term of the quadratic number pattert 2.2 Calculate the value of the $$ 48^{ ext {th }} $$ teno of the quadiatic number pattern 2.3 Show that the sum of the first differences of this quadratic number pattem can be given by $$ S_{n}=2 n^{2}+n $$

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The firs four terms of a quadratic number pattem are: $$ -1 ; 2 ; 9,20 $$ 2.1 Determine the general term of the quadratic number pattert 2.2 Calculate the value of the $$ 48^{	ext {th }} $$ teno of the quadiatic number pattern 2.3 Show that the sum of the first differences of this quadratic number pattem can be given by $$ S_{n}=2 n^{2}+n $$

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Alright, I need to solve this problem involving a quadratic number pattern. The first four terms are given as -1, 2, 9, and 20. Let's break down the problem into parts and tackle each one step by step.

**2.1 Determine the general term of the quadratic number pattern**

First, I need to find the general term of the quadratic sequence. A quadratic sequence has the form:
$$ a_n = an^2 + bn + c $$
where $$ a $$, $$ b $$, and $$ c $$ are constants to be determined.

Given the first four terms:
- $$ a_1 = -1 $$
- $$ a_2 = 2 $$
- $$ a_3 = 9 $$
- $$ a_4 = 20 $$

Let's set up equations based on these terms.

For $$ n = 1 $$:
$$ a(1)^2 + b(1) + c = -1 $$
$$ a + b + c = -1 $$  ...(1)

For $$ n = 2 $$:
$$ a(2)^2 + b(2) + c = 2 $$
$$ 4a + 2b + c = 2 $$  ...(2)

For $$ n = 3 $$:
$$ a(3)^2 + b(3) + c = 9 $$
$$ 9a + 3b + c = 9 $$  ...(3)

For $$ n = 4 $$:
$$ a(4)^2 + b(4) + c = 20 $$
$$ 16a + 4b + c = 20 $$  ...(4)

Now, I have a system of four equations with three unknowns. To solve for $$ a $$, $$ b $$, and $$ c $$, I can use methods like substitution or elimination. Let's try elimination.

Subtract equation (1) from equation (2):
$$ (4a + 2b + c) - (a + b + c) = 2 - (-1) $$
$$ 3a + b = 3 $$  ...(5)

Subtract equation (2) from equation (3):
$$ (9a + 3b + c) - (4a + 2b + c) = 9 - 2 $$
$$ 5a + b = 7 $$  ...(6)

Now, subtract equation (5) from equation (6):
$$ (5a + b) - (3a + b) = 7 - 3 $$
$$ 2a = 4 $$
$$ a = 2 $$

Now, substitute $$ a = 2 $$ back into equation (5):
$$ 3(2) + b = 3 $$
$$ 6 + b = 3 $$
$$ b = -3 $$

Finally, substitute $$ a = 2 $$ and $$ b = -3 $$ back into equation (1):
$$ 2 + (-3) + c = -1 $$
$$ -1 + c = -1 $$
$$ c = 0 $$

So, the general term of the quadratic sequence is:
$$ a_n = 2n^2 - 3n $$

**2.2 Calculate the value of the 48th term of the quadratic number pattern**

Using the general term $$ a_n = 2n^2 - 3n $$, let's find $$ a_{48} $$:
$$ a_{48} = 2(48)^2 - 3(48) $$
$$ a_{48} = 2(2304) - 144 $$
$$ a_{48} = 4608 - 144 $$
$$ a_{48} = 4464 $$

So, the 48th term is 4464.

**2.3 Show that the sum of the first differences of this quadratic number pattern can be given by $$ S_n = 2n^2 + n $$**

First, let's recall that the first difference of a sequence is the difference between consecutive terms. For a quadratic sequence, the first differences themselves form an arithmetic sequence.

Given the general term $$ a_n = 2n^2 - 3n $$, let's find the first difference $$ d_n = a_{n+1} - a_n $$:
$$ d_n = [2(n+1)^2 - 3(n+1)] - [2n^2 - 3n] $$
$$ d_n = [2(n^2 + 2n + 1) - 3n - 3] - [2n^2 - 3n] $$
$$ d_n = 2n^2 + 4n + 2 - 3n - 3 - 2n^2 + 3n $$
$$ d_n = (2n^2 - 2n^2) + (4n - 3n + 3n) + (2 - 3) $$
$$ d_n = 4n - 1 $$

So, the first difference is $$ d_n = 4n - 1 $$.

Now, the sum of the first $$ n $$ differences $$ S_n $$ is:
$$ S_n = \sum_{k=1}^{n} d_k = \sum_{k=1}^{n} (4k - 1) $$
$$ S_n = 4\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 $$
$$ S_n = 4\left(\frac{n(n+1)}{2}\right) - n $$
$$ S_n = 2n(n+1) - n $$
$$ S_n = 2n^2 + 2n - n $$
$$ S_n = 2n^2 + n $$

Therefore, the sum of the first differences of the quadratic number pattern is $$ S_n = 2n^2 + n $$.

1. The general term of the quadratic number pattern is $$ a_n = 2n^2 - 3n $$. 2. The 48th term of the quadratic number pattern is 4464. 3. The sum of the first differences of the quadratic number pattern is $$ S_n = 2n^2 + n $$.

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