1. \( \sum_{k=2}^{n}(5-2 k)=\frac{-800 n}{17} \) 2. \( \sum_{k=0}^{x} 108(2 / 3)^{r}=\frac{25200}{81} \)

It seems that the equations provided involve summations that cannot be directly solved using the tools available. Let's break down each equation step by step to find a solution. ### Problem 1: \[ \sum_{k=2}^{n}(5-2k) = \frac{-800n}{17} \] **Step 1: Calculate the summation.** The summation can be simplified as follows: \[ \sum_{k=2}^{n}(5-2k) = \sum_{k=2}^{n}5 - \sum_{k=2}^{n}2k \] Calculating each part: 1. The first part: \[ \sum_{k=2}^{n}5 = 5(n-1) \quad \text{(since there are } n-1 \text{ terms)} \] 2. The second part: \[ \sum_{k=2}^{n}2k = 2\sum_{k=2}^{n}k = 2\left(\frac{n(n+1)}{2} - 1\right) = n(n+1) - 2 \] Putting it all together: \[ \sum_{k=2}^{n}(5-2k) = 5(n-1) - (n(n+1) - 2) = 5n - 5 - n^2 - n + 2 = -n^2 + 4n - 3 \] **Step 2: Set the equation.** \[ -n^2 + 4n - 3 = \frac{-800n}{17} \] **Step 3: Clear the fraction by multiplying through by 17.** \[ -17n^2 + 68n - 51 = -800n \] \[ -17n^2 + 868n - 51 = 0 \] **Step 4: Solve the quadratic equation.** Now we can use the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -17, b = 868, c = -51 \). ### Problem 2: \[ \sum_{k=0}^{x} 108\left(\frac{2}{3}\right)^{k} = \frac{25200}{81} \] **Step 1: Recognize the summation as a geometric series.** The sum of a geometric series can be calculated using the formula: \[ S_n = a \frac{1 - r^{n+1}}{1 - r} \] where \( a = 108 \), \( r = \frac{2}{3} \), and \( n = x \). **Step 2: Set the equation.** \[ 108 \frac{1 - \left(\frac{2}{3}\right)^{x+1}}{1 - \frac{2}{3}} = \frac{25200}{81} \] This simplifies to: \[ 108 \cdot 3(1 - \left(\frac{2}{3}\right)^{x+1}) = \frac{25200}{81} \] \[ 324(1 - \left(\frac{2}{3}\right)^{x+1}) = \frac{25200}{81} \] **Step 3: Solve for \( x \).** Now we can isolate \( x \) and solve. Let's calculate the quadratic equation for the first problem and the geometric series for the second problem. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(-17n^{2}+868n-51=0\) - step1: Factor the expression: \(\left(-n+51\right)\left(17n-1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&-n+51=0\\&17n-1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&n=51\\&n=\frac{1}{17}\end{align}\) - step4: Rewrite: \(n_{1}=\frac{1}{17},n_{2}=51\) Solve the equation \( 324(1 - (\frac{2}{3})^{x+1}) = \frac{25200}{81} \). Solve the equation by following steps: - step0: Solve for \(x\): \(324\left(1-\left(\frac{2}{3}\right)^{x+1}\right)=\frac{25200}{81}\) - step1: Multiply the terms: \(324-324\left(\frac{2}{3}\right)^{x+1}=\frac{2800}{9}\) - step2: Move the expression to the left side: \(324-324\left(\frac{2}{3}\right)^{x+1}-\frac{2800}{9}=0\) - step3: Subtract the numbers: \(\frac{116}{9}-324\left(\frac{2}{3}\right)^{x+1}=0\) - step4: Rewrite the expression: \(-324\left(\frac{2}{3}\right)^{x+1}=-\frac{116}{9}\) - step5: Change the sign: \(324\left(\frac{2}{3}\right)^{x+1}=\frac{116}{9}\) - step6: Multiply by the reciprocal: \(324\left(\frac{2}{3}\right)^{x+1}\times \frac{1}{324}=\frac{116}{9}\times \frac{1}{324}\) - step7: Multiply: \(\left(\frac{2}{3}\right)^{x+1}=\frac{29}{729}\) - step8: Take the logarithm of both sides: \(\log_{\frac{2}{3}}{\left(\left(\frac{2}{3}\right)^{x+1}\right)}=\log_{\frac{2}{3}}{\left(\frac{29}{729}\right)}\) - step9: Evaluate the logarithm: \(x+1=\log_{\frac{2}{3}}{\left(\frac{29}{729}\right)}\) - step10: Move the constant to the right side: \(x=\log_{\frac{2}{3}}{\left(\frac{29}{729}\right)}-1\) - step11: Simplify: \(x=\log_{\frac{2}{3}}{\left(29\right)}-6\log_{\frac{2}{3}}{\left(3\right)}-1\) ### Solutions #### Problem 1: The equation we derived was: \[ -17n^2 + 868n - 51 = 0 \] The solutions for \( n \) are: \[ n_1 = \frac{1}{17}, \quad n_2 = 51 \] #### Problem 2: The equation we derived was: \[ 324(1 - (\\frac{2}{3})^{x+1}) = \\frac{25200}{81} \] The solution for \( x \) is: \[ x = \log_{\\frac{2}{3}}{(29)} - 6\log_{\\frac{2}{3}}{(3)} - 1 \] ### Summary of Results: 1. For the first problem, the values of \( n \) are \( \frac{1}{17} \) and \( 51 \). 2. For the second problem, \( x \) is expressed in logarithmic form as \( \log_{\\frac{2}{3}}{(29)} - 6\log_{\\frac{2}{3}}{(3)} - 1 \). If you need further simplification or numerical evaluation of \( x \), please let me know!