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Question #40706

Difficulty: 8/10 | 🟢 Free

The Question

Let Ak=(1k(k+1)2k(k+2)2(k+1)(k+3)1(k+2)(k+3))A_k = \begin{pmatrix} \frac{1}{k(k+1)} & \frac{2}{k(k+2)} \\ \frac{2}{(k+1)(k+3)} & \frac{1}{(k+2)(k+3)} \end{pmatrix} for k∈Nk \in \mathbb{N}. Evaluate det⁡(∑k=1∞Ak)\det \left( \sum_{k=1}^{\infty} A_k \right).

Options

  • A.

    −1112-\frac{11}{12}

  • B.

    −16-\frac{1}{6}

  • C.

    −512-\frac{5}{12}

  • D.

    112\frac{1}{12}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

V-N Method (Partial Fractions)|Determinant Product Rule|Inversion and Determinant Product

Why This Question Is Tricky

Common Mistakes

This question tests the V-N (telescopic) method for summation embedded within a matrix determinant.

The primary 'Lethal Collision' is the Linearity Trap: students often incorrectly assume that the determinant operator distributes over matrix addition, i.e., det⁡(∑Ak)=∑det⁡(Ak)\det(\sum A_k) = \sum \det(A_k).

If a student falls for this trap, they will compute det⁡(Ak)=−3k(k+1)(k+2)(k+3)\det(A_k) = \frac{-3}{k(k+1)(k+2)(k+3)} and sum it to get −1/6-1/6, which is provided as a distractor option.

The correct approach requires summing the matrices first using partial fractions, then taking the determinant of the resulting matrix, punishing the misconception that matrix summation commutes with the determinant function.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptV-N Method (Partial Fractions)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree