Question

Let \( f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix} \). If \( \sum_{n=1}^{k} f(n) = 98 \), then find \( k \).

Hint:

If only one row or column of a determinant contains the variable $n$, the summation sign can be taken inside that specific row or column.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
To find the sum of a determinant where only one column depends on the summation variable \( n \), we can apply the summation directly to the elements of that column while keeping the other columns constant.
Step 2: Key Formula or Approach:
1. \[ \sum_{n=1}^{k} f(n) = \begin{vmatrix} \sum n & -1 & -5 \\ \sum (-2n^2) & 3(2k+1) & 2k+1 \\ \sum (-3n^3) & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix} \] 2. Use standard formulas: \[ \sum n = \frac{k(k+1)}{2}, \quad \sum n^2 = \frac{k(k+1)(2k+1)}{6}, \quad \sum n^3 = \frac{k^2(k+1)^2}{4} \]
Step 3: Detailed Explanation:
After substituting: \[ = \begin{vmatrix} \frac{k(k+1)}{2} & -1 & -5 \\ -\frac{k(k+1)(2k+1)}{3} & 3(2k+1) & 2k+1 \\ -\frac{3k^2(k+1)^2}{4} & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix} \] Now simplifying the determinant (by expansion or standard reduction), it reduces to: \[ k(k+1) \] So, \[ \sum_{n=1}^{k} f(n) = k(k+1) \] Given: \[ k(k+1) = 98 \] \[ k^2 + k - 98 = 0 \] \[ (k+14)(k-7) = 0 \] Step 4: Final Answer:
\[ k = 7 \]