JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If \(f: \mathbf{N} \rightarrow \mathbf{Z}\) is defined by
\(f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}\), \(k \in \mathbf{N}\),
and \(\sum_{n=1}^{k} f(n) = 98\), then \(k\) is equal to :
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
Let \(S = \displaystyle\sum_{n=1}^{k} f(n)\), where \(f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}\) Since only the first column depends on \(n\), the summation can be taken inside that column:…
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