Given the geometric series: \( \frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\ldots=\mathrm{K} \) 2.2.1 \( \quad \) Determine the value of \( K \) if the series has 21 terms. 2.2.2 \( \quad \) Determine the largest value of \( n \) for which \( T_{n}>\frac{5}{8192} \)

To tackle the geometric series \( \frac{5}{8} + \frac{5}{16} + \frac{5}{32} + \ldots \), we first identify the first term \( a = \frac{5}{8} \) and the common ratio \( r = \frac{1}{2} \) (since each term is half of the previous one). **2.2.1** To find the sum \( K \) of the first 21 terms of the series, we use the formula for the sum of the first \( n \) terms of a geometric series, given by \( S_n = a \frac{1 - r^n}{1 - r} \): \[ K = \frac{5}{8} \cdot \frac{1 - \left(\frac{1}{2}\right)^{21}}{1 - \frac{1}{2}} = \frac{5}{8} \cdot \frac{1 - \frac{1}{2^{21}}}{\frac{1}{2}} = \frac{5}{8} \cdot \frac{2 \left(1 - \frac{1}{2^{21}}\right)}{1} = \frac{5}{4} \left(1 - \frac{1}{2^{21}}\right). \] Calculating this further gives: \[ K = \frac{5}{4} - \frac{5}{4 \cdot 2^{21}} = \frac{5}{4} - \frac{5}{8388608} \approx 1.25 - 0.000000596046 \approx 1.249999404. \] **2.2.2** To determine the largest value of \( n \) for which \( T_n > \frac{5}{8192} \), we first express the \( n \)-th term of the series: \[ T_n = ar^{n-1} = \frac{5}{8} \left(\frac{1}{2}\right)^{n-1} = \frac{5}{8 \cdot 2^{n-1}}. \] Setting up the inequality: \[ \frac{5}{8 \cdot 2^{n-1}} > \frac{5}{8192}. \] Cancelling 5 from both sides gives: \[ \frac{1}{8 \cdot 2^{n-1}} > \frac{1}{8192}. \] Cross-multiplying yields: \[ 8192 > 8 \cdot 2^{n-1}. \] On simplifying, we find: \[ 1024 > 2^{n-1}. \] Recognizing \( 1024 = 2^{10} \), we therefore have: \[ 2^{10} > 2^{n-1} \implies 10 > n - 1 \implies n < 11. \] Thus, the largest value of \( n \) satisfying this is \( n = 10 \). So, the results are: - \( K \approx 1.249999404 \) for the sum. - The largest \( n \) for \( T_n > \frac{5}{8192} \) is \( n = 10 \).