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Odd And Even Coefficient Sums

Mediummathematics

If the expansion of (1 + x)^n is split into terms with even index and terms with odd index, the sum of the even-indexed coefficients separately equals which value?

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About This Question

Subject
mathematics
Chapter
binomial theorem and its simple applications
Topic
odd and even coefficient sums
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drilleven-odd-coefficientspaired-substitutionsubset-countingbinomial-identity

Solution

Correct Answer:

Separating even and odd indexed coefficients is achieved by combining the substitutions x = 1 and x = -1, a paired-evaluation technique that JEE Advanced uses to split sums. Let E = C(n,0) + C(n,2) + C(n,4) + ... be the even-index sum and O = C(n,1) + C(n,3) + ... be the odd-index sum. Substituting x = 1 in (1 + x)^n gives E + O = 2^n, while substituting x = -1 gives E - O = 0 since (1 - 1)^n = 0 for positive n. Adding these two equations yields 2E = 2^n, so E = 2^{n-1}; subtracting yields the same for O. Hence the even-indexed coefficient sum is 2^{n-1}. Option 2^n is the total of all coefficients, not just the even-indexed half. Option 2^{n+1} doubles the full sum erroneously. Option n^2 has no connection to the binomial structure. The symmetry E = O = 2^{n-1} reflects the equal numbers of even-sized and odd-sized subsets of an n-element set. Plausibility check: for n = 3, the even-index coefficients 1 + 3 = 4 = 2^{2}, exactly 2^{n-1}, confirming the result.

This medium difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of odd and even coefficient sums. It appeared in the 2025 exam.

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