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Question #41499

Difficulty: 8/10 | 🟢 Free

The Question

Let CnC_n be the number of distinct terms in the multinomial expansion of (x1+x2+x3+x4)n(x_1 + x_2 + x_3 + x_4)^n. Evaluate the sum S=∑n=1201CnS = \sum_{n=1}^{20} \frac{1}{C_n}

Options

  • A.

    125253\frac{125}{253}

  • B.

    1251518\frac{125}{1518}

  • C.

    5051518\frac{505}{1518}

  • D.

    250253\frac{250}{253}

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

Partial Fractions Decomposition|Multinomial Theorem|Combinatorial Counting

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Combinatorics (Multinomial Theorem) and Algebra (Telescoping Series via Partial Fractions).

The primary trap lies in the combinatorial counting phase: students frequently treat identical items as distinct, omitting the 3!3! division and incorrectly evaluating Cn=(n+1)(n+2)(n+3)C_n = (n+1)(n+2)(n+3), which leads to the incorrect denominator and option B.

A secondary trap is the partial fraction decomposition of 1/(x(x+1)(x+2))1/(x(x+1)(x+2)), where students often forget the 1/21/2 scaling factor, leading directly to option D.

The problem requires precise execution across multiple distinct mathematical domains to reach the correct finite limit.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptPartial Fractions Decomposition
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree