Given the quadratic number pattern: $$ -3 ; 6 ; 27 ; 60 ; \ldots $$ 2.2.1 Determine the general term of the pattern in the form $$ T_{n}=a n^{2}+b n+c $$. $$ 2.2 .2 \quad \begin{array}{l} ext { Calculate the value of the } 50^{ ext {th }} ext { term of the pattern. } \ 2.2 .3\end{array} \begin{array}{l} ext { Show that the sum of the first } n ext { first-differences of this pattern can be } \ ext { given by } S_{n}=6 n^{2}+3 n .\end{array} $$ $$ 2.2 .4 \quad \begin{array}{l} ext { How many consecutive first-differences were added to the first term of } \ ext { the quadratic number pattern to obtain a term in the quadratic number } \ ext { pattern that has a value of } 21060 ext { ? }\end{array} $$ ION 3 If $$ \sum_{k= ho}^{\infty} 4.3^{2-k}=\frac{2}{9} $$, determine the value of $$ p $$.

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Given the quadratic number pattern: $$ -3 ; 6 ; 27 ; 60 ; \ldots $$ 2.2.1 Determine the general term of the pattern in the form $$ T_{n}=a n^{2}+b n+c $$. $$ 2.2 .2 \quad \begin{array}{l}	ext { Calculate the value of the } 50^{	ext {th }} 	ext { term of the pattern. } \ 2.2 .3\end{array} \begin{array}{l}	ext { Show that the sum of the first } n 	ext { first-differences of this pattern can be } \ 	ext { given by } S_{n}=6 n^{2}+3 n .\end{array} $$ $$ 2.2 .4 \quad \begin{array}{l}	ext { How many consecutive first-differences were added to the first term of } \ 	ext { the quadratic number pattern to obtain a term in the quadratic number } \ 	ext { pattern that has a value of } 21060 	ext { ? }\end{array} $$ ION 3 If $$ \sum_{k=ho}^{\infty} 4.3^{2-k}=\frac{2}{9} $$, determine the value of $$ p $$.

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Let's solve the problem step by step.

### 2.2.1 Determine the general term of the pattern in the form $$ T_{n}=a n^{2}+b n+c $$.

We have the sequence: $$ -3, 6, 27, 60, \ldots $$

Let's denote the terms as follows:
- $$ T_1 = -3 $$
- $$ T_2 = 6 $$
- $$ T_3 = 27 $$
- $$ T_4 = 60 $$

We need to find $$ a, b, c $$ such that:
$$
T_n = a n^2 + b n + c
$$

We can set up a system of equations using the known terms:

1. For $$ n = 1 $$:
   $$
   a(1^2) + b(1) + c = -3 \quad \Rightarrow \quad a + b + c = -3 \quad \text{(1)}
   $$

2. For $$ n = 2 $$:
   $$
   a(2^2) + b(2) + c = 6 \quad \Rightarrow \quad 4a + 2b + c = 6 \quad \text{(2)}
   $$

3. For $$ n = 3 $$:
   $$
   a(3^2) + b(3) + c = 27 \quad \Rightarrow \quad 9a + 3b + c = 27 \quad \text{(3)}
   $$

Now we have the following system of equations:
$$
\begin{align*}
1. & \quad a + b + c = -3 \
2. & \quad 4a + 2b + c = 6 \
3. & \quad 9a + 3b + c = 27 \
\end{align*}
$$

Next, we can eliminate $$ c $$ by subtracting equation (1) from equations (2) and (3):

From (2) - (1):
$$
(4a + 2b + c) - (a + b + c) = 6 - (-3) \
3a + b = 9 \quad \text{(4)}
$$

From (3) - (1):
$$
(9a + 3b + c) - (a + b + c) = 27 - (-3) \
8a + 2b = 30 \quad \Rightarrow \quad 4a + b = 15 \quad \text{(5)}
$$

Now we have a new system of equations:
$$
\begin{align*}
4. & \quad 3a + b = 9 \
5. & \quad 4a + b = 15 \
\end{align*}
$$

Subtract equation (4) from (5):
$$
(4a + b) - (3a + b) = 15 - 9 \
a = 6
$$

Substituting $$ a = 6 $$ back into equation (4):
$$
3(6) + b = 9 \
18 + b = 9 \
b = 9 - 18 = -9
$$

Now substituting $$ a $$ and $$ b $$ back into equation (1):
$$
6 - 9 + c = -3 \
c = -3 + 3 = 0
$$

Thus, the general term is:
$$
T_n = 6n^2 - 9n
$$

### 2.2.2 Calculate the value of the $$ 50^{\text{th}} $$ term of the pattern.

Now we can calculate $$ T_{50} $$:
$$
T_{50} = 6(50^2) - 9(50)
$$

Calculating $$ 50^2 $$:
$$
50^2 = 2500
$$

Now substituting:
$$
T_{50} = 6(2500) - 9(50) = 15000 - 450 = 14550
$$

### 2.2.3 Show that the sum of the first $$ n $$ first-differences of this pattern can be given by $$ S_n = 6n^2 + 3n $$.

The first differences $$ D_n $$ can be calculated as:
$$
D_n = T_n - T_{n-1}
$$

Calculating $$ D_n $$:
$$
D_n = (6n^2 - 9n) - (6(n-1)^2 - 9(n-1))
$$
Expanding $$ T_{n-1} $$:
$$
T_{n-1} = 6(n^2 - 2n + 1) - 9(n - 1) = 6n^2 - 12n + 6 - 9n + 9 = 6n^2 - 21n + 15
$$
Thus,
$$
D_n = (6n^2 - 9n) - (6n^2 - 21n + 15) = 12n - 15
$$

Now, the sum of the first $$ n $$ first-differences:
$$
S_n = \sum_{k=1}^{n} D_k = \sum_{k=1}^{n} (12k - 15)
$$
This can be simplified:
$$
S_n = 12\sum_{k=1}^{n} k - 15n = 12\left(\frac{n(n+1)}{2}\right) - 15n = 6n(n+1) - 15n
$$
$$
= 6n^2 + 6n - 15n = 6n^2 - 9n
$$

### 2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of $$ 21060 $$?

We need to find $$ n $$ such that:
$$
T_n = 21060
$$
Using the general term:
$$
6n^2 - 9n = 21060
$$
Rearranging gives:
$$
6n^2 - 9n - 21060 = 0
$$
Dividing the entire equation by 3:
$$
2n^2 - 3n - 7020 = 0
$$

Now we can solve this quadratic equation using the quadratic formula:
$$
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-7020)}}{2(2)}
$$
Calculating the discriminant:
\[
b^2 - 4ac = 9 + 4 \cdot 
The general term of the pattern is $$ T_n = 6n^2 - 9n $$. The 50th term is 14,550. The sum of the first $$ n $$ first-differences is $$ S_n = 6n^2 - 9n $$. To reach a term of 21,060, 20 consecutive first-differences were added to the first term.

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