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Question #41095

Difficulty: 8/10 | 🟢 Free

The Question

Let $C_r = \binom{20}{r}$ for $0 \le r \le 20$ . Define a sequence $f(k) = \sum_{r=0}^k (-1)^r C_r$ . Let $A = \sum_{k=0}^{10} |f(k)|$ and $B = \sum_{k=0}^{10} f(k)$ . The value of $A - |B|$ is:

Options

A.
$2^{18} + \binom{18}{9}$
B.
$2^{19} + \binom{18}{9}$
C.
$2^{18} + \binom{19}{9}$
D.
$2^{19} - \binom{18}{9}$

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

Symmetry and Invariance|Properties of Binomial Coefficients|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize the telescoping nature of alternating binomial sums and carefully handle the 'Modulus Trap'.

Students often attempt to expand the double summation for $B$ or misapply continuous approximations, failing to realize that $f(k)$ simplifies to a single binomial coefficient $(-1)^k \binom{19}{k}$ .

The modulus in $A$ strips the alternating sign, forcing the evaluation of a half-sum of binomial coefficients requiring symmetry $\binom{19}{k} = \binom{19}{19-k}$ .

Meanwhile, $B$ retains the alternating nature, telescoping once more.

A common trap is ignoring the modulus or assuming the alternating sum $B$ evaluates to zero, which leads directly to the distractor $2^{18} + \binom{19}{9}$ .

Successfully navigating the parity constraints allows for the final elegant simplification via Pascal's identity.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Symmetry and Invariance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free