4.2 Given the geometric series: \( 54+18+6+\cdots+54\left(\frac{1}{3}\right)^{n-1} \) 4.2.1 Show that the sum of the first \( n \) terms is given by \( 81-81\left(\frac{1}{3}\right)^{n} \). 4.2.2 Calculate the smallest value of \( n \) for which the Sum of the first \( n \) terms is greater than 80,99 . 4.2.3 What does \( \left(\frac{1}{3}\right)^{n} \) approach as \( n \rightarrow \infty \) ? 4.2.4 Determine the value of \( \sum_{k=1}^{\infty} 54\left(\frac{1}{3}\right)^{k-1} \), using QUESTION 4.2.1 and 4.2 .3 or in another way.

**4.2.1** The sum of the first \( n \) terms of a geometric series with first term \( a \) and common ratio \( r \) is given by \[ S_n = a\frac{1-r^n}{1-r}. \] Here, \( a = 54 \) and \( r = \frac{1}{3} \). Therefore, \[ S_n = 54\frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}} = 54\frac{1-\left(\frac{1}{3}\right)^n}{\frac{2}{3}}. \] Multiplying numerator and denominator by 3 gives \[ S_n = 54\cdot\frac{3\left[1-\left(\frac{1}{3}\right)^n\right]}{2} = 81\left[1-\left(\frac{1}{3}\right)^n\right]. \] Thus, \[ S_n = 81 - 81\left(\frac{1}{3}\right)^n. \] --- **4.2.2** We need to find the smallest \( n \) such that \[ S_n > 80.99. \] Using the expression from 4.2.1, \[ 81 - 81\left(\frac{1}{3}\right)^n > 80.99. \] Subtract \( 81 \) from both sides: \[ -81\left(\frac{1}{3}\right)^n > -0.01. \] Multiplying both sides by \(-1\) (which reverses the inequality): \[ 81\left(\frac{1}{3}\right)^n < 0.01. \] Divide both sides by 81: \[ \left(\frac{1}{3}\right)^n < \frac{0.01}{81} = \frac{1}{8100}. \] Taking the natural logarithm of both sides, \[ \ln\left(\left(\frac{1}{3}\right)^n\right) < \ln\left(\frac{1}{8100}\right). \] This simplifies to \[ n\ln\left(\frac{1}{3}\right) < \ln\left(\frac{1}{8100}\right). \] Since \(\ln\left(\frac{1}{3}\right) < 0\), dividing by it reverses the inequality: \[ n > \frac{\ln\left(\frac{1}{8100}\right)}{\ln\left(\frac{1}{3}\right)}. \] Notice that \[ \ln\left(\frac{1}{8100}\right) = -\ln(8100) \quad \text{and} \quad \ln\left(\frac{1}{3}\right) = -\ln(3). \] Thus, \[ n > \frac{-\ln(8100)}{-\ln(3)} = \frac{\ln(8100)}{\ln(3)}. \] We can write \(8100\) as \(81 \times 100\). Since \(81 = 3^4\), \[ \ln(8100) = \ln(81) + \ln(100) = 4\ln(3) + \ln(100). \] Therefore, \[ n > \frac{4\ln(3) + \ln(100)}{\ln(3)} = 4 + \frac{\ln(100)}{\ln(3)}. \] Now, using approximate values \(\ln(100) \approx 4.60517\) and \(\ln(3) \approx 1.09861\), \[ \frac{\ln(100)}{\ln(3)} \approx \frac{4.60517}{1.09861} \approx 4.191. \] Thus, \[ n > 4 + 4.191 \approx 8.191. \] Since \( n \) must be an integer, the smallest value is \( n = 9 \). --- **4.2.3** As \( n \rightarrow \infty \), \[ \left(\frac{1}{3}\right)^n \rightarrow 0. \] --- **4.2.4** We want to find \[ \sum_{k=1}^{\infty} 54\left(\frac{1}{3}\right)^{k-1}. \] For an infinite geometric series with \( |r| < 1 \), the sum is \[ S_{\infty} = \frac{a}{1-r}. \] Here, \( a = 54 \) and \( r = \frac{1}{3} \): \[ S_{\infty} = \frac{54}{1-\frac{1}{3}} = \frac{54}{\frac{2}{3}} = 54 \times \frac{3}{2} = 81. \] Alternatively, using the result from 4.2.1 and taking the limit as \( n\to\infty \): \[ \lim_{n\to\infty}\left[81-81\left(\frac{1}{3}\right)^n\right] = 81 - 81\cdot0 = 81. \] The final answers are: - 4.2.1: \( S_n = 81 - 81\left(\frac{1}{3}\right)^n \) - 4.2.2: The smallest \( n \) is \( 9 \). - 4.2.3: \( \left(\frac{1}{3}\right)^n \rightarrow 0 \) as \( n \rightarrow \infty \). - 4.2.4: \( \displaystyle \sum_{k=1}^{\infty} 54\left(\frac{1}{3}\right)^{k-1} = 81 \).