Calculate the value of \( n \) if it is given that: \( \sum_{k=2}^{n}(5-2 k)=-\frac{800 n}{17} \)
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To solve the equation \( \sum_{k=2}^{n}(5-2 k)=-\frac{800 n}{17} \), we start by calculating the left-hand side of the equation, which is the summation of the expression \( (5 - 2k) \). 1. **Calculate the summation**: \[ S = \sum_{k=2}^{n} (5 - 2k) = \sum_{k=2}^{n} 5 - \sum_{k=2}^{n} 2k \] The first term can be simplified: \[ \sum_{k=2}^{n} 5 = 5(n - 1) \] because there are \( n - 1 \) terms from \( k=2 \) to \( k=n \). For the second term: \[ \sum_{k=2}^{n} 2k = 2\sum_{k=2}^{n} k = 2\left(\frac{n(n+1)}{2} - 1\right) = n(n+1) - 2 \] Thus we have: \[ S = 5(n - 1) - (n(n + 1) - 2) = 5n - 5 - n(n + 1) + 2 = 5n - n^2 - n - 5 + 2 = -n^2 + 4n - 3 \] 2. **Set up the equation**: Now we equate this to the right-hand side: \[ -n^2 + 4n - 3 = -\frac{800n}{17} \] 3. **Multiply through by 17 to eliminate the fraction**: \[ -17n^2 + 68n - 51 = -800n \] Rearranging gives: \[ -17n^2 + 68n + 800n - 51 = 0 \implies -17n^2 + 868n - 51 = 0 \] We can multiply through by -1 for convenience: \[ 17n^2 - 868n + 51 = 0 \] 4. **Use the quadratic formula**: The quadratic formula is given by: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here \( a = 17 \), \( b = -868 \), and \( c = 51 \). Calculate the discriminant: \[ b^2 - 4ac = (-868)^2 - 4 \times 17 \times 51 = 752624 - 3468 = 749156 \] Finding the square root of the discriminant: \[ \sqrt{749156} \approx 866.6 \] Now plug into the quadratic formula: \[ n = \frac{868 \pm 866.6}{34} \] This results in two possible values: - \( n_1 \approx \frac{1734.6}{34} \approx 51 \) - \( n_2 \approx \frac{1.4}{34} \approx 0.04 \) (not valid since \( n \) must be an integer and greater than or equal to 2) Thus, the only viable solution is: \[ \boxed{51} \]