Evaluating a polynomial at 1 adds its coefficients. Evaluating at -1 changes the signs of its odd-power terms. Combining those two evaluations can isolate a whole group of coefficients at once.
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Define the two sums
In this original example, let E be the sum of C(7,k) over even k, and O the corresponding sum over odd k. The binomial expansion of (1 + x)7 contains exactly these coefficients.
Substituting x = 1 gives E + O = 27 = 128. Substituting x = -1 gives E - O = 0. The second equality uses the signs of the powers, not the signs of the positive coefficients themselves.
Combine the equations
Adding the two equations gives 2E = 128, so E = 64. Subtracting them gives O = 64 as well.
For a direct numerical check, the even-index sum is C(7,0) + C(7,2) + C(7,4) + C(7,6) = 1 + 21 + 35 + 7 = 64. The evaluation method obtains that total without separately computing the four coefficients.
Explain the equality by pairing subsets
Take a set of seven labelled objects and choose one distinguished object. Pair every subset not containing it with the subset obtained by adding it. Each pair has one even-sized subset and one odd-sized subset.
Every subset appears in exactly one pair. Since there are 128 subsets altogether, each parity class has 64. This is an independent counting explanation of the same identity.
State the range of the general result
For a positive integer n, the same reasoning gives equal even- and odd-index binomial sums, each 2n-1. The empty-set case n = 0 needs separate treatment: its only subset is even-sized, so the two totals are not equal.
This is an original AMC 12 learning exercise, not an official problem. When using evaluations to filter terms, state which terms keep or reverse their signs and why the resulting equations determine the requested sum.
Continue reading: AMC 12 Algebra: Find One Binomial Coefficient Without Expanding Everything.
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Official information and further reading: Source 1
