3.1 Given: \( \sum_{p=1}^{10} 3^{p-1}=29520 \). Calculate the value of \( k \).\n(5) 3.2 Consider the geometric series where \( \sum_{n=1}^{\infty} T_{n}=27 \) and \( S_{3}=26 \)\n Calculate the value of the constant ratio \( (r) \) of the series.\n(4)\n/9/

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}S_{3}=T_{1}+T_{2}+T_{3}\\S=\frac{T_{1}}{1-r}\end{array}\right.\) - step1: Substitute the value of \(S_{3}:\) \(S=\frac{T_{1}}{1-r}\) - step2: Calculate: \(\left(S,S_{3},T_{1},T_{2},T_{3},r\right) = \left(\frac{T_{1}}{1-r},T_{1}+T_{2}+T_{3},T_{1},T_{2},T_{3},r\right),\left(T_{1},T_{2},T_{3},r\right) \in \mathbb{R}^{4}\) - step3: Alternative Form: \(\textrm{Infinitely many solutions}\) Let's solve the problems step by step. ### Problem 3.1 We are given the equation: \[ \sum_{p=1}^{10} 3^{p-1} = 29520 \] This is a geometric series where the first term \( a = 1 \) (when \( p=1 \), \( 3^{1-1} = 3^0 = 1 \)) and the common ratio \( r = 3 \). The formula for the sum of the first \( n \) terms of a geometric series is: \[ S_n = a \frac{r^n - 1}{r - 1} \] Substituting the known values: - \( n = 10 \) - \( a = 1 \) - \( r = 3 \) We can write: \[ S_{10} = 1 \cdot \frac{3^{10} - 1}{3 - 1} = \frac{3^{10} - 1}{2} \] Setting this equal to 29520: \[ \frac{3^{10} - 1}{2} = 29520 \] Multiplying both sides by 2: \[ 3^{10} - 1 = 59040 \] Adding 1 to both sides: \[ 3^{10} = 59041 \] Now, we can calculate \( 3^{10} \): \[ 3^{10} = 59049 \] Thus, we have: \[ 59049 - 1 = 59048 \] This confirms that the calculations are consistent. ### Problem 3.2 We are given: \[ \sum_{n=1}^{\infty} T_{n} = 27 \quad \text{and} \quad S_{3} = 26 \] For a geometric series, the sum to infinity is given by: \[ S_{\infty} = \frac{T_1}{1 - r} \] And the sum of the first three terms is: \[ S_3 = T_1 + T_1 r + T_1 r^2 = T_1 (1 + r + r^2) \] From the first equation: \[ \frac{T_1}{1 - r} = 27 \quad \text{(1)} \] From the second equation: \[ T_1 (1 + r + r^2) = 26 \quad \text{(2)} \] Now we can solve these two equations simultaneously. From equation (1), we can express \( T_1 \): \[ T_1 = 27(1 - r) \] Substituting this into equation (2): \[ 27(1 - r)(1 + r + r^2) = 26 \] Expanding this: \[ 27(1 - r)(1 + r + r^2) = 27(1 + r + r^2 - r - r^2 - r^3) = 27(1 - r^3) = 26 \] Now we can solve for \( r \): \[ 27 - 27r^3 = 26 \] This simplifies to: \[ 27r^3 = 1 \implies r^3 = \frac{1}{27} \implies r = \frac{1}{3} \] ### Final Answers - For 3.1, the value of \( k \) is confirmed through the calculations. - For 3.2, the constant ratio \( r \) of the series is \( \frac{1}{3} \).