(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 .

To tackle the geometric series \( 3 + 9 + 27 + \ldots \), we identify the first term \( a = 3 \) and the common ratio \( r = 3 \). (1) The formula for the sum of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] For \( S_8 \): \[ S_8 = 3 \frac{3^8 - 1}{3 - 1} = 3 \frac{6561 - 1}{2} = 3 \frac{6560}{2} = 3 \times 3280 = 9840 \] (2) To find \( k \) such that \( S_k = 88572 \): Set the sum formula equal to 88572: \[ 3 \frac{3^k - 1}{3 - 1} = 88572 \implies \frac{3^k - 1}{2} = 29524 \implies 3^k - 1 = 59048 \implies 3^k = 59049 \] Taking the logarithm, we find \( k = 10 \) since \( 3^{10} = 59049 \). (3) To find the least number of terms for which the sum will be greater than 20000, we set up the inequality: \[ 3 \frac{3^n - 1}{2} > 20000 \] Solving gives: \[ 3^n - 1 > \frac{40000}{3} \approx 13333 \implies 3^n > 13334 \] Now, we find \( n \): Start estimating powers of 3: - \( 3^8 = 6561 \) (not enough) - \( 3^9 = 19683 \) (sufficient) Thus, the least number of terms for which the sum exceeds 20000 is \( n = 9 \).