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Sum Of Binomial Coefficients

Easymathematics

The combinatorial sum C(n,0) + C(n,1) + C(n,2) + ... + C(n,n), formed from a complete row of binomial coefficients, evaluates to which closed-form expression?

Select the correct option:

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About This Question

Subject
mathematics
Chapter
binomial theorem and its simple applications
Topic
sum of binomial coefficients
Difficulty
Easy
Year
2025
Tags
advanced-calculus-drillsum-of-binomial-coefficientsidentity-proofsubstitutionpascal-row

Solution

Correct Answer:

A foundational identity in the binomial theorem states that the sum of all coefficients in a row of Pascal's triangle equals 2^n, derived by a clever choice of variable, which JEE Advanced uses constantly. Start from the expansion (1 + x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n. Substituting x = 1 makes every power of x equal to one, so the left side becomes (1 + 1)^n = 2^n and the right side becomes exactly the required sum of coefficients. Hence the sum equals 2^n. Option 2^{n-1} represents only the sum of the even-indexed or odd-indexed coefficients, which is half the total. Option n^2 has no basis in the binomial structure. Option n! confuses arrangements with the coefficient sum. The substitution x = 1 is the canonical proof technique here. Plausibility check: for a small case n = 3, the coefficients 1, 3, 3, 1 sum to 8 = 2^3, confirming the formula and the doubling behaviour as n increases by one.

This easy difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of sum of binomial coefficients. It appeared in the 2025 exam.

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