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Question #40709

Difficulty: 8/10 | 🟢 Free

The Question

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three non-coplanar vectors such that no two are orthogonal. A sequence of vectors is defined as v⃗k=xk(a⃗×(b⃗×c⃗))+yk(b⃗×(c⃗×a⃗))+zk(c⃗×(a⃗×b⃗))\vec{v}_k = x_k (\vec{a} \times (\vec{b} \times \vec{c})) + y_k (\vec{b} \times (\vec{c} \times \vec{a})) + z_k (\vec{c} \times (\vec{a} \times \vec{b})) where xk=4(k2+k−1)(k+2)!x_k = \frac{4(k^2+k-1)}{(k+2)!}, yk=2k(k−1)(k+1)!y_k = \frac{2^k(k-1)}{(k+1)!}, and zk=λk(k+1)(k+2)z_k = \frac{\lambda}{k(k+1)(k+2)}. If ∑k=1∞v⃗k=0⃗\sum_{k=1}^\infty \vec{v}_k = \vec{0}, then the value of λ\lambda is

Options

  • A.

    0

  • B.

    4

  • C.

    8

  • D.

    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)|Vector Triple Product Expansion|Linear Independence vs Dependence

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of the Method of Differences (V-N method) and Vector Algebra.

The primary trap is assuming the vector triple products A⃗,B⃗,C⃗\vec{A}, \vec{B}, \vec{C} are linearly independent, which would lead students to incorrectly set X=0,Y=0,Z=0X=0, Y=0, Z=0 and conclude no solution exists or λ=0\lambda=0.

Recognizing the Jacobi Identity (A⃗+B⃗+C⃗=0⃗\vec{A}+\vec{B}+\vec{C}=\vec{0}) is mandatory to reduce the system to a linearly independent basis, revealing the hidden constraint X=Y=ZX=Y=Z.

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