Alternating Sum Of Coefficients
Evaluating the alternating combination C(n,0) - C(n,1) + C(n,2) - C(n,3) + ... over a full binomial row produces which value when n is a positive integer?
Select the correct option:
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About This Question
- Subject
- mathematics
- Chapter
- binomial theorem and its simple applications
- Topic
- alternating sum of coefficients
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
0
The alternating sum of binomial coefficients is evaluated by substituting a negative value into the standard expansion, a method JEE Advanced expects students to recognise instantly. Begin with (1 + x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n and put x = -1. The left side becomes (1 - 1)^n = 0^n, which equals 0 for every positive integer n. The right side becomes C(n,0) - C(n,1) + C(n,2) - C(n,3) + ..., exactly the alternating sum in question. Therefore the alternating sum is 0. Option 1 ignores that the base 1 - 1 vanishes. Option 2^n is the result of substituting x = +1, not x = -1, so it answers a different question. Option (-1)^n incorrectly treats the base as -1 raised to n rather than (1 - 1)^n. This zero result is equivalent to saying the number of even-sized subsets equals the number of odd-sized subsets. Plausibility check: for n = 2 the coefficients give 1 - 2 + 1 = 0, and for n = 3 they give 1 - 3 + 3 - 1 = 0, confirming the cancellation in general.
This medium difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of alternating sum of coefficients. It appeared in the 2025 exam.
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