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Question #41639

Difficulty: 8/10 | 🟢 Free

The Question

Let PnP_n be a family of planes given by the equation x+y+z=dnx + y + z = d_n, where dnd_n is the coefficient of t2t^2 in the polynomial (1+t)n+3−(1+t)n+1(1+t)^{n+3} - (1+t)^{n+1}. A line LL is given by the symmetric equations x−20231=y−20241=z−20251\frac{x - 2023}{1} = \frac{y - 2024}{1} = \frac{z - 2025}{1}. Let Qn(xn,yn,zn)Q_n(x_n, y_n, z_n) be the unique point of intersection of the plane PnP_n and the line LL. Find the value of ∑n=18(xn+yn+zn)\sum_{n=1}^{8} (x_n + y_n + z_n).

Options

  • A.

    96

  • B.

    83

  • C.

    54

  • D.

    100

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Parametric form pitfalls|Variable plane intercepts|The Telescopic Series Trap|Binomial Coefficient Summation Properties

Why This Question Is Tricky

Common Mistakes

This question tests students on two distinct JEE Advanced traps.

First, the 'Red Herring Trap': students often waste time parameterizing the line LL and solving for the intersection point, failing to realize that the sum of the coordinates is directly given by the plane's equation.

Second, the 'Telescopic Series Index Shift Trap': the summation ∑(An+2−An)\sum (A_{n+2} - A_n) has an index shift of 2, meaning two terms survive at both the upper and lower bounds (A10+A9−A2−A1A_{10} + A_9 - A_2 - A_1).

Students blindly applying the standard f(n)−f(n−1)f(n) - f(n-1) telescopic property will incorrectly cancel terms, leading to distractors like 54 (computing A10−A1A_{10} - A_1) or 100 (forgetting the lower bounds).

The value 83 serves as a secondary distractor for arithmetic miscalculations.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptParametric form pitfalls
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree