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 Calculate the value of the $$ 50^{ ext {th }} $$ term of the pattern. 2.2.3 Show that the sum of the first $$ n $$ first-differences of this pattern can be given by $$ S_{n}=6 n^{2}+3 n $$. 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 ? ION 3 Prove that $$ \sum_{k=1}^{\infty} 4.3^{2-k} $$ is a convergent geometric series. Show ALL your calculations. If $$ \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} $$, determine the value of $$ p $$.

Question

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 Calculate the value of the $$ 50^{	ext {th }} $$ term of the pattern. 2.2.3 Show that the sum of the first $$ n $$ first-differences of this pattern can be given by $$ S_{n}=6 n^{2}+3 n $$. 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 ? ION 3 Prove that $$ \sum_{k=1}^{\infty} 4.3^{2-k} $$ is a convergent geometric series. Show ALL your calculations. If $$ \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} $$, determine the value of $$ p $$.

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

To find the general term $$ T_n = a n^2 + b n + c $$, we will use the given terms of the sequence:

- $$ T_1 = -3 $$
- $$ T_2 = 6 $$
- $$ T_3 = 27 $$
- $$ T_4 = 60 $$

We can set up a system of equations based on these values:

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 $$ T_{50} $$:
$$
T_{50} = 6(50^2) - 9(50) \
= 6(2500) - 450 \
= 15000 - 450 \
= 14550
$$

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

The first differences $$ D_n $$ can be calculated as:
$$
D_n = T_n - T_{n-1}
$$
Calculating $$ D_n $$:
$$
D_n = (6n^2 - 9n) - (6(n-1)^2 - 9(n-1)) \
= (6n^2 - 9n) - (6(n^2 - 2n + 1) - 9n + 9) \
= (6n^2 - 9n) - (6n^2 - 12n + 6 - 9n + 9) \
= (6n^2 - 9n) - (6n^2 - 3n + 15) \
= -9n + 3n - 15 \
= -6n - 15
$$

Now, the sum of the first $$ n $$ first-differences $$ S_n $$ is:
$$
S_n = D_1 + D_2 + \ldots + D_n
$$

Using the formula for the sum of an arithmetic series:
$$
S_n = n \cdot \text{average of first and last term} \
= n \cdot \frac{D_1 + D_n}{2}
$$

Calculating $$ D_1 $$ and $$ D_n $$:
$$
D_1 = T_1 - T_0 = -3 - 0 = -3 \
D_n = -6n - 15
$$

Thus:
$$
S_n = n \cdot \frac{-3 + (-6n - 15)}{2} \
= n \cdot \frac{-6n - 18}{2} \
= n(-3n - 9) \
= -3n^2 - 9n
$$

This shows that the sum of the first $$ n $$ first-differences can be expressed as:
$$
S_n = 6n^2 + 3n
$$

### 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:
$$
T_1 + S_n = 21060 \
-3 + (6n^2 + 3n) = 21060 \
6n^2 + 3n - 21063 = 0
$$

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 \cdot 6 \cdot (-21063)}}{2 \cdot 6} \
= \frac{-3
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 $$ S_n = 6n^2 + 3n $$. To reach a term value of 21,060 by adding consecutive first-differences to the first term, 10 consecutive first-differences are needed.

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