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Question #40688

Difficulty: 8/10 | 🟢 Free

The Question

Let RR be the term independent of xx in the binomial expansion of (x−1x2)9\left( x - \frac{1}{x^2} \right)^9. Let SnS_n denote the sum of the first nn terms of an Arithmetic-Geometric Progression whose kk-th term is given by tk=(2k+1)(R84)k−1t_k = (2k+1) \left( \frac{R}{84} \right)^{k-1}. Find the value of S57+S84S_{57} + S_{84}.

Options

  • A.

    -25

  • B.

    10587

  • C.

    -27

  • D.

    145

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

Arithmetic-Geometric Progression (AGP)|General Term Formulation|Term Independent of x|Sign Flipping Rule

Why This Question Is Tricky

Common Mistakes

This question tests the integration of binomial expansion with sequence summation, specifically targeting the 'Sign Flipping Rule'.

The trap lies in the negative binomial coefficient (−1)k(-1)^k when k=3k=3, which makes the AGP common ratio r=−1r = -1 rather than 11.

Students who ignore the negative sign will incorrectly sum a standard arithmetic progression (leading to the distractor 10587).

Furthermore, the alternating AGP creates a split-case logical branch where the sum formula oscillates depending on the parity of the index nn.

Evaluating S57S_{57} and S84S_{84} forces the student to correctly identify and apply both the odd and even parity cases of the sequence sum.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptArithmetic-Geometric Progression (AGP)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree