(d) Consider the geometric series \( 3+9+27+\ldots \) (1) Calculate \( S_{8} \). (2) Determine the value of \( k \) if \( S_{k}=88572 \). (3) Determine the least number of terms for which the sum will be greater than 20000.

The geometric series given is \( 3 + 9 + 27 + \ldots \) with a first term \( a = 3 \) and a common ratio \( r = 3 \). (1) To calculate \( S_{8} \), we use the formula for the sum of the first \( n \) terms of a geometric series: \[ S_n = a \frac{r^n - 1}{r - 1} \] Substituting \( a = 3 \), \( r = 3 \), and \( n = 8 \): \[ S_8 = 3 \frac{3^8 - 1}{3 - 1} = 3 \frac{6561 - 1}{2} = 3 \frac{6560}{2} = 3 \times 3280 = 9840 \] So, \( S_{8} = 9840 \). (2) We need to find \( k \) such that \( S_k = 88572 \): \[ S_k = 3 \frac{3^k - 1}{3 - 1} = 3 \frac{3^k - 1}{2} = \frac{3^{k+1} - 3}{2} \] Setting this equal to 88572: \[ \frac{3^{k+1} - 3}{2} = 88572 \] Multiplying both sides by 2: \[ 3^{k+1} - 3 = 177144 \] Adding 3 to both sides: \[ 3^{k+1} = 177147 \] Now we can express 177147 as a power of 3. We find that \( 177147 = 3^{11} \) (since \( 3^{11} = 177147 \)). Setting the exponents equal gives us: \[ k + 1 = 11 \Rightarrow k = 10 \] Thus, the value of \( k \) is \( 10 \). (3) We want the sum to be greater than 20000: \[ S_n = 3 \frac{3^n - 1}{3 - 1} = \frac{3^{n+1} - 3}{2} > 20000 \] Multiplying both sides by 2: \[ 3^{n+1} - 3 > 40000 \] Adding 3 to both sides: \[ 3^{n+1} > 40003 \] Taking logarithm base 3: \[ n + 1 > \log_3(40003) \] Calculating \( \log_3(40003) \): Using the change of base formula: \[ \log_3(40003) = \frac{\log_{10}(40003)}{\log_{10}(3)} \approx \frac{4.602}{0.477} \approx 9.64 \] Thus, \[ n + 1 > 9.64 \Rightarrow n > 8.64 \] The least integer satisfying this is \( n = 9 \). Therefore, at least \( 9 \) terms are needed for the sum to exceed \( 20000 \).