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: \( 25\left(\frac{1-t}{3}\right)+5\left(\frac{1-1}{3}\right)^{2}+\left(\frac{1-1}{3}\right)^{3}+ \). \( \qquad \) 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 \( 70^{\text {th }} \) term. Leave your answer in the form \( a . b^{\circ} \) where \( a, b \) and \( p \) at integers.

To unravel the mystery of the quadratic sequence \( 6 ; x ; 26 ; 45 ; y ; \ldots \), let’s find \( x \) and \( y \)! The differences between terms are \( x - 6 \), \( 26 - x \), and \( 45 - 26 = 19 \), leading us to the second differences, which should be constant for a quadratic sequence. With a little math magic, we find \( x = 13 \) and calculating for \( y \) gives us \( y = 66 \). Now, addressing the series \( 220 + 213 + 206 + \ldots - 11 \), it’s an arithmetic series with a first term \( a = 220 \) and a common difference \( d = -7 \). The last term, -11, leads us to calculate the number of terms as \( n = \frac{(220 - (-11))}{7} + 1 = 33 \). The sum can be found using \( S_n = \frac{n}{2} (a + l) = \frac{33}{2} (220 - 11) = 3465 \). For sigma notation, it's beautifully expressed as \( \sum_{k=0}^{32} (220 - 7k) \). And for that bouncy batt? With each bounce losing 10% of its height, we demonstrate it rises smaller until the summation of an infinite geometric series converges, staying under that 290 m limit. It’s all about clever sequences and numerals dancing together!