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:
When summing a determinant where only one column (or row) contains terms in \(n\), the summation sign can be shifted inside to that specific column.
Step 2: Key Formula or Approach:
The sum is \(\sum_{n=1}^k f(n) = \begin{vmatrix} \sum n & -1 & -5
\sum -2n^2 & 3(2k + 1) & 2k + 1
\sum -3n^3 & 3k(2k + 1) & 3k(k + 2) + 1 \end{vmatrix}\).
Using standard sum formulas:
\(\sum_{n=1}^k n = \frac{k(k+1)}{2}\)
\(\sum_{n=1}^k n^2 = \frac{k(k+1)(2k+1)}{6}\)
\(\sum_{n=1}^k n^3 = \frac{k^2(k+1)^2}{4}\)
Step 3: Detailed Explanation:
Substitute the sums into the first column:
\[ \begin{vmatrix} \frac{k(k+1)}{2} & -1 & -5
\frac{-2k(k+1)(2k+1)}{6} & 3(2k + 1) & 2k + 1
\frac{-3k^2(k+1)^2}{4} & 3k(2k + 1) & 3k(k + 2) + 1 \end{vmatrix} = 98 \]
Observe the common factors in the first column and take them out.
After applying column transformations \(C_2 \rightarrow C_2 + C_1\) and \(C_3 \rightarrow C_3 + 5C_1\) and simplifying:
\[ \frac{k(k+1)(2k+1)}{6} \cdot 7 = 98 \Rightarrow k(k+1)(2k+1) = 84 \]
By trial and error (testing small values for \(k\)):
For \(k = 3 \Rightarrow 3(4)(7) = 84\).
Step 4: Final Answer:
The value of \(k\) is 3.