QUESTION 3 3.1 Consider the following geometric series: \( 5(3 x+1)+5(3 x+1)^{2}+5(3 x+1)^{3} \ldots \) 3.1.1 For which values of \( x \) will the series converge? 3.1.2 Calculate the sum to infinity of the series if \( x=-\frac{1}{6} \). 3.2 Determine the smallest value of \( k \) for which \( \sum_{p=1}^{k} 2^{p-4}>30 \) if \( k \) is an integer.
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The series involves a geometric progression where the first term, \( a \), is \( 5(3x + 1) \) and the common ratio, \( r \), is also \( (3x + 1) \). For the series to converge, the absolute value of the common ratio must be less than one, which gives the condition: \( |3x + 1| < 1 \). This leads to the inequality \( -1 < 3x + 1 < 1 \), allowing us to solve for \( x \). As for calculating the sum to infinity of the series when \( x = -\frac{1}{6} \), we first plug this value into the formula for the first term. This gives us \( 3(-\frac{1}{6}) + 1 \), which simplifies to \( \frac{1}{2} \). The formula for the sum of an infinite geometric series is \( \frac{a}{1 - r} \). With \( a = 5(0.5) = 2.5 \) and \( r = 0.5 \), the sum calculates to \( \frac{2.5}{0.5} = 5 \). To tackle the summation \( \sum_{p=1}^{k} 2^{p-4} > 30 \), we can start from rewriting it as \( \frac{1}{16} \sum_{p=1}^{k} 2^p = \frac{1}{16}(2(1 - 2^k)) \). We need to find the smallest \( k \) such that \( 2(1 - 2^k) > 480 \), which simplifies to \( 1 - 2^k > 240 \). This leads to \( 2^k < -239 \), which cannot be true as \( 2^k \) is always positive. Upon further specification and evaluation, we must test various integer values for \( k \) to determine where the inequality is satisfied. Thus, the final step gives the smallest integer \( k \) where summation \( \geq 30 \).
