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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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: $$ 6 ; x ; 26 ; 45 ; y ; \ldots $$ In a quadratic sequence, the second differences are constant. 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: - $$ d_1 = a_2 - a_1 = x - 6 $$ - $$ d_2 = a_3 - a_2 = 26 - x $$ - $$ d_3 = a_4 - a_3 = 45 - 26 = 19 $$ - $$ d_4 = a_5 - a_4 = y - 45 $$ Now, we calculate the second differences: - $$ d_2 - d_1 = (26 - x) - (x - 6) = 32 - 2x $$ - $$ d_3 - d_2 = 19 - (26 - x) = x - 7 $$ - $$ d_4 - d_3 = (y - 45) - 19 = y - 64 $$ Setting the second differences equal to each other: 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. Series: $$ 220 + 213 + 206 + \ldots - 11 $$ This is an arithmetic series where: - First term $$ a = 220 $$ - Common difference $$ d = 213 - 220 = -7 $$ #### 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 Write the series in sigma-notation. The series can be expressed in sigma notation as: $$ \sum_{k=0}^{33} (220 - 7k) $$ ### 2.3 Total Distance of the Bouncing Ball The ball is dropped from a height of $$ 15 $$ m and loses $$ 10\% $$ of its height on each bounce. The height after each bounce is: - First bounce: $$ 15 \times 0.9 = 13.5 $$ - Second bounce: $$ 13.5 \times 0.9 = 12.15 $$ - And so on... The total distance $$ D $$ can be calculated as: $$ D = 15 + 2 \sum_{n=1}^{\infty} 15 \times (0.9)^n $$ The series $$ \sum_{n=1}^{\infty} (0.9)^n $$ is a geometric series with first term $$ a = 0.9 $$ and ratio $$ r = 0.9 $$: $$ \sum_{n=1}^{\infty} (0.9)^n = \frac{0.9}{1 - 0.9} = 9 $$ Thus, $$ D = 15 + 2 \times 15 \times 9 = 15 + 270 = 285 $$ Since $$ 285 < 290 $$, the total distance cannot exceed $$ 290 $$ m. ### 2.4 Series Convergence #### 2.4.1 For which value(s) of $$ l $$ will the series converge? The series converges if the common ratio $$ r < 1 $$. The common ratio is $$ \frac{1 - l}{3} $$: $$ \frac{1 - l}{3} < 1 \implies 1 - l < 3 \implies l > -2 $$ #### 2.4.2 If $$ t = 15 $$, calculate the sum to infinity of the series if it exists. Substituting $$ t = 15 $$: $$ S = 25 \left(\frac{1 - 15}{3}\right) + 5 \left(\frac{1 - 1}{3}\right)^{2} + \left(\frac{1 - 1}{3}\right)^{3} + \ldots $$ The first term dominates, and since the other terms vanish, we focus on: $$ S = 25 \left(\frac{-14}{3}\right) = -\frac{350}{3} $$ ### 2.5 Determine the $$ 70^{\text{th}} $$ term Given $$ S_{\infty} = 2^{x-1} + 3 $$, the $$ n^{\text{th}} $$ term $$ T_n $$ can be expressed as: $$ T_n = S_n - S_{n-1} $$ Assuming $$ S_n = 2^{n-1} + 3 $$: $$ S_{n-1} = 2^{n-2} + 3 $$ Thus, $$ T_n = (2^{n-1} + 3) - (2^{n-2} + 3) = 2^{n-1} - 2^{n-2} = 2^{n-2} $$ 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: - $$ x = 13 $$, $$ y = 70 \ - $$ x = 13 $$, $$ y = 70 $$ - The sum of the series is 3553. - The series in sigma-notation is $$ \sum_{k=0}^{33} (220 - 7k) $$. - The total distance the ball will bounce is 285 m, which is less than 290 m. - The series converges for $$ l > -2 $$. - When $$ t = 15 $$, the sum to infinity is $$ -\frac{350}{3} $$. - The 70th term is $$ 2^{68} $$.

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