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Question #40689

Difficulty: 8/10 | 🟢 Free

The Question

For $k \ge 2$ , let $C_k$ be the coefficient of $x y z^{k-2}$ in the multinomial expansion of $(x+y+z)^k$ , and let $C_1 = 0$ . For $k \ge 1$ , let $D_k$ be the coefficient of $x y^{k-1}$ in the binomial expansion of $(x+y)^k$ . A sequence is defined by $T_k = (C_k + D_k + 1) k!$ . If $S_n = \sum_{k=1}^n T_k$ , find the maximum integer $m$ such that $5^m$ divides $S_{125}$ .

Options

A.
30
B.
31
C.
34
D.
35

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

Method of Differences|Multinomial Theorem|Exponent of Prime in n!

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of the Multinomial Theorem, Method of Differences, and Number Theory (Legendre's Formula).

The primary trap lies in the telescoping sum's result: $S_n = n(n+1)!$ .

Students frequently make an off-by-one index error during telescoping, yielding $(n-1)n!$ , or they incorrectly assume the sum collapses simply to $(n+1)!$ .

For $n=125$ , both of these flawed sums yield an exponent of 31 for the prime 5.

The lethal collision occurs because students must recognize that the coefficient $n=125$ itself contributes additional powers of 5 ( $5^3$ ), making the correct exponent $31 + 3 = 34$ .

This forces a shift from algebraic sequence manipulation to careful prime factorization.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Method of Differences
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free