ESTION 2 Consider the following quadratic sequence: $$ 6 ; x ; 26 ; 45 ; y ; \ldots $$ Determine the values of $$ x $$ and $$ y $$. 2 Given the following series: $$ 220+213+206+\ldots-11 $$ 2.2.1 Calculate the sum of the series. 2.2.2 Write the series in sigma-notation. 2.3 A batt is dropped from a beight of 15 m . It bounces back and loses $$ 10 \% $$ of its prev height on each bounce. Show that the total distance the ball will bounce cannot exi 290 m . 2.4 Given: $$ 25\left(\frac{1-t}{3} ight)+5\left(\frac{1-1}{3} ight)^{2}+\left(\frac{1-1}{3} ight)^{3}+ $$. $$ \qquad $$ 2.4.1 For which value(s) of $$ l $$ will the series converge? 2.4.2 If $$ t=15 $$, calculate the sum to infinity of the series if it exists. 2.5 The sum of the first $$ n $$ terms of a sequence is $$ S_{\infty}=2^{x-1}+3 $$. Deternine the $$ 70^{ ext {th }} $$ term. Leave your answer in the form $$ a . b^{\circ} $$ where $$ a, b $$ and $$ p $$ at integers.

Question

ESTION 2 Consider the following quadratic sequence: $$ 6 ; x ; 26 ; 45 ; y ; \ldots $$ Determine the values of $$ x $$ and $$ y $$. 2 Given the following series: $$ 220+213+206+\ldots-11 $$ 2.2.1 Calculate the sum of the series. 2.2.2 Write the series in sigma-notation. 2.3 A batt is dropped from a beight of 15 m . It bounces back and loses $$ 10 \% $$ of its prev height on each bounce. Show that the total distance the ball will bounce cannot exi 290 m . 2.4 Given: $$ 25\left(\frac{1-t}{3}ight)+5\left(\frac{1-1}{3}ight)^{2}+\left(\frac{1-1}{3}ight)^{3}+ $$. $$ \qquad $$ 2.4.1 For which value(s) of $$ l $$ will the series converge? 2.4.2 If $$ t=15 $$, calculate the sum to infinity of the series if it exists. 2.5 The sum of the first $$ n $$ terms of a sequence is $$ S_{\infty}=2^{x-1}+3 $$. Deternine the $$ 70^{	ext {th }} $$ term. Leave your answer in the form $$ a . b^{\circ} $$ where $$ a, b $$ and $$ p $$ at integers.

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Let's break down the problem step by step.

### 1. Quadratic Sequence: $$ 6 ; x ; 26 ; 45 ; y ; \ldots $$

A quadratic sequence has a constant second difference. Let's denote the terms as follows:
- $$ a_1 = 6 $$
- $$ a_2 = x $$
- $$ a_3 = 26 $$
- $$ a_4 = 45 $$
- $$ a_5 = y $$

First, we calculate the first differences:
- $$ a_2 - a_1 = x - 6 $$
- $$ a_3 - a_2 = 26 - x $$
- $$ a_4 - a_3 = 45 - 26 = 19 $$
- $$ a_5 - a_4 = y - 45 $$

Now, we calculate the second differences:
- Second difference between first differences:
  - $$ (26 - x) - (x - 6) = 26 - 2x + 6 = 32 - 2x $$
  - $$ 19 - (26 - x) = 19 - 26 + x = x - 7 $$
  - $$ (y - 45) - 19 = y - 64 $$

Setting the second differences equal:
1. $$ 32 - 2x = x - 7 $$
2. $$ x - 7 = y - 64 $$

Now, let's solve these equations.

#### Solving the first equation:
$$
32 - 2x = x - 7
$$
$$
32 + 7 = 3x
$$
$$
39 = 3x \implies x = 13
$$

#### Solving the second equation:
$$
13 - 7 = y - 64
$$
$$
6 = y - 64 \implies y = 70
$$

Thus, the values are:
- $$ x = 13 $$
- $$ y = 70 $$

### 2. Given the series: $$ 220 + 213 + 206 + \ldots - 11 $$

This is an arithmetic series where:
- First term $$ a = 220 $$
- Common difference $$ d = 213 - 220 = -7 $$

#### 2.2.1 Calculate the sum of the series.

To find the number of terms $$ n $$:
$$
a_n = a + (n-1)d
$$
Setting $$ a_n = -11 $$:
$$
-11 = 220 + (n-1)(-7)
$$
$$
-11 - 220 = -7(n-1)
$$
$$
-231 = -7(n-1) \implies n - 1 = \frac{231}{7} = 33 \implies n = 34
$$

Now, we can calculate the sum $$ S_n $$:
$$
S_n = \frac{n}{2} (a + a_n) = \frac{34}{2} (220 - 11) = 17 \times 209 = 3553
$$

#### 2.2.2 Write the series in sigma-notation.

The series can be expressed in sigma notation as:
$$
\sum_{k=0}^{33} (220 - 7k)
$$

### 3. A ball dropped from a height of 15 m

The ball bounces back to $$ 90\% $$ of its previous height after each bounce. The heights of the bounces form a geometric series.

The heights of the bounces are:
- First drop: $$ 15 $$
- First bounce: $$ 15 \times 0.9 = 13.5 $$
- Second bounce: $$ 13.5 \times 0.9 = 12.15 $$
- And so on...

The total distance $$ D $$ is:
$$
D = 15 + 2(13.5 + 12.15 + \ldots)
$$

The series $$ 13.5 + 12.15 + \ldots $$ is a geometric series with:
- First term $$ a = 13.5 $$
- Common ratio $$ r = 0.9 $$

The sum of an infinite geometric series is given by:
$$
S = \frac{a}{1 - r} = \frac{13.5}{1 - 0.9} = \frac{13.5}{0.1} = 135
$$

Thus, the total distance is:
$$
D = 15 + 2 \times 135 = 15 + 270 = 285 \text{ m}
$$

### 4. Given series: $$ 25\left(\frac{1-t}{3}\right)+5\left(\frac{1-1}{3}\right)^{2}+\left(\frac{1-1}{3}\right)^{3}+ $$

This series appears to be a geometric series. Let's analyze it.

#### 4.4.1 For which value(s) of $$ t $$ will the series converge?

The series converges if the common ratio $$ r < 1 $$. The common ratio is:
$$
r = \frac{1 - t}{3}
$$
Setting $$ |r| < 1 $$:
$$
\left|\frac{1 - t}{3}\right| < 1 \implies -3 < 1 - t < 3
$$
This gives:
$$
-2 < -t < 4 \implies -4 < t < 2
$$

#### 4.4.2 If $$ t = 15 $$, calculate the sum to infinity of the series if it exists.

Since $$ t = 15 $$ is outside the convergence range, the series does not converge.

### 5. The sum of the first $$ n $$ terms of a sequence is $$ S_{\infty} = 2^{x-1} + 3 $$.

To find the $$ 70^{th} $$ term, we need to express the $$ n^{th} $$ term $$ T_n $$:
$$
T_n = S_n - S_{n-1}
$$
Assuming $$ S_n = 2^{n-1} + 3 $$:
$$
S_{n-1} = 2^{(n-1)-1} + 3 = 2^{n-2} + 3
$$
Thus,
$$
T_n = (2^{n-1} + 3) - (2^{n-2} + 3) = 2^{n-1} - 2^{n-2} = 2^{n-2}
$$

Now, for $$ n = 70 $$:
$$
T_{70} = 2^{70-2} = 2^{68}
$$

The answer in the form $$ a \cdot b^p $$ is:
$$
1 \cdot 2^{68}
$$

### Summary of Answers
- $$ x = 13 $$ and $$ y = 70 $$ - The sum of the series is 3553. - The series in sigma-notation is $$ \sum_{k=0}^{33} (220 - 7k) $$. - The total distance the ball will bounce is 285 m. - The series converges for $$ -4 < t < 2 $$. For $$ t = 15 $$, the series does not converge. - The 70th term is $$ 2^{68} $$.

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