The sum of the even-power coefficients in a binomial expansion can be found without writing every term. Evaluate the polynomial at x = 1 and x = −1: adding the two results cancels odd powers, while subtracting cancels even powers.
For P(x) = (1 + x)^n with positive integer n, both the even-power coefficient sum and the odd-power coefficient sum equal 2^(n−1). For a more general example, P(x) = (2 + x)^4 has even-power coefficient sum (3⁴ + 1⁴)/2 = 41 and odd-power sum (3⁴ − 1⁴)/2 = 40. This guide explains the identity, handles signed coefficients, parameter questions and weighted sums, and shows why checking the exact quantity requested matters more than brute-force expansion.
Understand what an even-power coefficient means
In a polynomial P(x) = c₀ + c₁x + c₂x² + c₃x³ + …, the even-power coefficients are c₀, c₂, c₄ and so on. The constant term counts because x⁰ has even exponent zero.
The odd-power coefficients are c₁, c₃, c₅ and so on. Each coefficient includes its sign. A negative x³ term contributes a negative coefficient to the odd-power sum.
A common error is to confuse “even coefficients” with coefficients whose numerical values are even integers. Here “even” describes the exponent of x, not the parity of the coefficient value.
For example, in P(x) = 1 + 5x + 7x² − 3x³, the even-power coefficient sum is 1 + 7 = 8, while the odd-power sum is 5 − 3 = 2. Reading the language correctly is the first essential step.
Why evaluating at x = 1 sums all coefficients
For any polynomial P(x) = c₀ + c₁x + c₂x² + … + c_nx^n, substitute x = 1. Every power 1^r equals 1.
Therefore P(1) = c₀ + c₁ + c₂ + … + c_n, which is the sum of all coefficients including their signs.
For P(x) = (2 + x)^4, P(1) = (2 + 1)^4 = 81. The entire coefficient sum is 81 without expanding the fourth power.
This method works for general polynomials, not only binomials. Evaluating at a convenient input can reveal a sum of coefficients much faster than listing terms one by one.
Why evaluating at x = −1 separates alternating signs
Now substitute x = −1. Even powers of −1 equal +1, while odd powers equal −1.
Thus P(−1) = c₀ − c₁ + c₂ − c₃ + c₄ − … . It is the even-power coefficient sum minus the odd-power coefficient sum.
For P(x) = (2 + x)^4, P(−1) = (2 − 1)^4 = 1. The alternating coefficient sum is 1.
The two evaluations P(1) and P(−1) provide two linear equations in the two unknown sums. Solving them gives the even and odd sums separately without finding individual coefficients.
Derive the even–odd separation formulas
Let E denote the sum of even-power coefficients and O denote the sum of odd-power coefficients, including their actual signs.
Evaluating at x = 1 gives P(1) = E + O. Evaluating at x = −1 gives P(−1) = E − O.
Add these equations to obtain 2E = P(1) + P(−1), so E = [P(1) + P(−1)]/2.
Subtract the second from the first to obtain 2O = P(1) − P(−1), so O = [P(1) − P(−1)]/2. These identities are exact for every polynomial and should be understood as simple simultaneous equations rather than isolated facts to memorise.
A complete worked example with (2 + x) to the fourth power
Find the even-power and odd-power coefficient sums of P(x) = (2 + x)^4. Calculate P(1) = 3⁴ = 81 and P(−1) = 1⁴ = 1.
The even-power sum is E = (81 + 1)/2 = 41. The odd-power sum is O = (81 − 1)/2 = 40.
No binomial coefficient table is needed. The answer is obtained from two short evaluations and one addition or subtraction.
The result is plausible because E + O = 41 + 40 = 81, matching the total coefficient sum. Also E − O = 1, matching the alternating sum. These checks ensure the two requested quantities were not accidentally exchanged.
Verify the full expansion once
The ordinary binomial expansion is (2 + x)^4 = 16 + 32x + 24x² + 8x³ + x⁴.
Its even-power coefficients are 16, 24 and 1. Their sum is 16 + 24 + 1 = 41.
Its odd-power coefficients are 32 and 8. Their sum is 32 + 8 = 40.
The direct calculation agrees with the evaluation method. This example demonstrates why the latter is particularly valuable for larger exponents: the work remains almost unchanged even when the expansion contains many terms.
Why the two sums of (1 + x)^n are equal
Consider P(x) = (1 + x)^n for a positive integer n. Evaluating at 1 gives P(1) = 2^n.
At x = −1, P(−1) = (1 − 1)^n = 0 for n ≥ 1. Therefore E = [2^n + 0]/2 = 2^(n−1).
Similarly O = [2^n − 0]/2 = 2^(n−1). The even and odd coefficient sums are equal.
This identity follows from binomial symmetry and cancellation, not from each even-indexed coefficient matching an odd-indexed coefficient individually. The total sums agree even though the individual values may be different.
Check with n = 5
For (1 + x)^5, the coefficients are 1, 5, 10, 10, 5 and 1, corresponding to powers x⁰ through x⁵.
The even-power coefficients are 1, 10 and 5. Their sum is 16. The odd-power coefficients are 5, 10 and 1, also summing to 16.
The general formula gives 2^(5−1) = 16, confirming the result. The total coefficient sum is 32, equal to (1 + 1)^5.
This is a useful small check, but the identity is not dependent on testing n = 5. The x = ±1 derivation establishes it for every positive integer n.
The special case n = 0 must be separated
When n = 0, (1 + x)^0 = 1. The only coefficient is the constant 1.
Therefore the even-power sum is E = 1 and the odd-power sum is O = 0. They are not both 2^(−1).
The derivation E = O = 2^(n−1) used (1 − 1)^n = 0. At n = 0, the polynomial is simply 1, so that zero-evaluation step is not valid in the same way.
This edge case is a reminder to read the assumed range of n. Many school questions specify that n is a positive integer, which is exactly the condition needed for the usual equal-sums identity.
A general binomial has unequal even and odd sums
For P(x) = (a + bx)^n, evaluation at x = 1 gives (a + b)^n, while evaluation at x = −1 gives (a − b)^n.
Thus E = [(a + b)^n + (a − b)^n]/2 and O = [(a + b)^n − (a − b)^n]/2.
The two sums are equal only when P(−1) = 0, which depends on the specific coefficients and exponent. For positive integer n, a = b is one common case.
This general formula extends the simple (1 + x)^n identity without requiring any new binomial expansion. It is useful when a and b are non-unit numbers and the individual coefficients would become large.
Example with the stronger x coefficient two
Let P(x) = (1 + 2x)^5. Then P(1) = 3⁵ = 243.
At x = −1, P(−1) = (1 − 2)^5 = (−1)^5 = −1. Therefore E = (243 − 1)/2 = 121 and O = (243 + 1)/2 = 122.
The odd-power sum is one larger than the even-power sum in this example. This does not contradict the equal-sum identity for (1 + x)^n, because the binomial’s x coefficient is now 2 instead of 1.
A quick check is E + O = 243 and E − O = −1. Both recover the two polynomial evaluations used to derive the sums.
Check the coefficients in the fifth-power example
The expansion (1 + 2x)^5 has coefficients 1, 10, 40, 80, 80 and 32.
The even-power sum is 1 + 40 + 80 = 121, while the odd-power sum is 10 + 80 + 32 = 122.
The direct sums agree with the x = ±1 method. This is a helpful demonstration that even when the coefficients are all positive, the even and odd groups need not have the same total.
The equality observed for (1 + x)^n comes from the special cancellation at x = −1, not from the mere fact that all binomial coefficients are positive.
A negative x coefficient changes the signed odd sum
Now take Q(x) = (2 − x)^4. At x = 1, Q(1) = 1⁴ = 1. At x = −1, Q(−1) = 3⁴ = 81.
Therefore E = (1 + 81)/2 = 41, while O = (1 − 81)/2 = −40.
Indeed the expansion is 16 − 32x + 24x² − 8x³ + x⁴. The even-power coefficients sum to 16 + 24 + 1 = 41, and the odd-power coefficients sum to −32 − 8 = −40.
The odd sum is negative because coefficients include their signs. If the question asks for the sum of absolute values of odd coefficients, that is a different task and would give +40 here.
Distinguish signed sums from sums of magnitudes
The wording “sum of coefficients of odd powers” normally includes their signs. For (2 − x)^4, that signed sum is −40.
But the sum of magnitudes of its odd coefficients is |−32| + |−8| = 40. The two quantities differ by sign.
The x = ±1 method gives signed sums directly. It does not automatically give a sum of absolute values when coefficient signs vary in a more complicated pattern.
Read whether the question specifies coefficients, magnitudes or absolute values before applying the formula. A mathematically correct evaluation can still answer the wrong question if the requested quantity was misread.
An unknown exponent can be recovered from the even sum
Suppose n is a positive integer and the sum of even-power coefficients in (1 + x)^n is 32. The identity gives 2^(n−1) = 32.
Since 32 = 2⁵, we have n − 1 = 5, so n = 6.
Check: (1 + x)^6 has total coefficient sum 64 and zero alternating sum, so its even and odd sums are both 32.
This is a parameter-recovery application of the identity. Instead of calculating individual binomial coefficients and solving a large expression in n, the student uses the evaluation relationship to reduce the problem to a simple index equation.
A parameter inside the binomial can be determined too
Suppose P(x) = (2 + bx)^4 and its even-power coefficient sum is 41 while its odd-power sum is 40.
Then P(1) = E + O = 81, so (2 + b)^4 = 81. Also P(−1) = E − O = 1, so (2 − b)^4 = 1.
The real solutions of the first relation are 2 + b = ±3, giving b = 1 or −5. The second relation requires 2 − b = ±1, giving b = 1 or 3.
The intersection is b = 1. Both supplied coefficient sums are needed to identify the admissible parameter, and it is important not to discard negative fourth-root possibilities without checking their intersection.
Why setting x = 0 does not separate parity sums
Evaluating P(0) gives the constant coefficient c₀ and removes every positive-power term. It is useful for finding the constant term, not the sum of all even-power coefficients.
For (2 + x)^4, P(0) = 16. But the even-power sum is 41, because it also includes the x² and x⁴ coefficients.
The special points x = 1 and x = −1 work because their integer powers follow simple, predictable patterns: every power of 1 is 1, and powers of −1 alternate signs.
Choosing the correct evaluation input is a method-selection skill. A student should know what each simple substitution isolates before using it.
The even part of any polynomial
A polynomial P(x) can be split into its even-power component E(x) and odd-power component O(x).
The formulas are E(x) = [P(x) + P(−x)]/2 and O(x) = [P(x) − P(−x)]/2. Adding removes odd-power terms, while subtracting removes even-power terms.
For P(x) = 1 + 5x + 7x² − 3x³, the even part is 1 + 7x² and the odd part is 5x − 3x³.
Evaluating E(1) and O(1) gives precisely the coefficient sums 8 and 2. This broader algebraic structure explains why the x = ±1 trick works for every polynomial and not only for binomial powers.
An extension using differentiation and weighted sums
A related identity is the weighted sum Σ rC(n, r) for (1 + x)^n. Differentiating gives n(1 + x)^(n−1) = Σ rC(n, r)x^(r−1).
Set x = 1 to obtain Σ rC(n, r) = n2^(n−1). The weights r arise naturally because differentiation brings the exponent down as a multiplier.
For n ≥ 2, the even-indexed and odd-indexed weighted sums are equal. Differentiating at x = −1 gives zero, which establishes the alternating weighted sum is zero. Each group is therefore n2^(n−2).
This is enrichment rather than a requirement for every introductory binomial exercise. It shows how evaluating a polynomial and its derivative at carefully chosen inputs can reveal different patterns of coefficients.
A weighted-sum check at n = 4
For (1 + x)^4, the coefficients are 1, 4, 6, 4 and 1.
The even-indexed weighted sum is 0·1 + 2·6 + 4·1 = 16. The odd-indexed weighted sum is 1·4 + 3·4 = 16.
The formula n2^(n−2) with n = 4 gives 4·2² = 16, agreeing with both groups. Their total is 32, equal to n2^(n−1) = 4·8.
This provides a numerical check of the derivative extension and reinforces the distinction between an ordinary coefficient sum and one that multiplies each coefficient by its power index.
An alternating sum is a direct x = −1 evaluation
For the ordinary binomial coefficients, the alternating sum C(n,0) − C(n,1) + C(n,2) − … equals (1 − 1)^n.
For positive integer n, this is zero. It is not necessary to pair terms manually or expand each combination number.
For a general polynomial P, the alternating signed coefficient sum is P(−1). For (2 + x)^4 it equals 1, while for (1 + 2x)^5 it equals −1.
These examples show why one universal alternating-sum answer does not apply to every binomial with different constants. The correct evaluation is based on the actual expression given in the question.
What happens if the expression has only even powers
Consider P(x) = (1 + x²)^5. Every term produced by this expansion has an even power of x because each selected x² contributes a multiple of two to the exponent.
Therefore its odd-power coefficient sum is zero. Its even-power coefficient sum equals the total coefficient sum P(1) = 2⁵ = 32.
The x = ±1 method agrees: P(1) = (1 + 1)^5 = 32 and P(−1) = (1 + (−1)²)^5 = 32. Thus E = 32 and O = 0.
This example cautions against applying E = O = 2^(n−1) to every expression that contains a binomial. The special equality belongs to (1 + x)^n, not automatically to (1 + x²)^n.
A short exam method
First identify whether the task asks for all coefficients, even-power coefficients, odd-power coefficients or an alternating sum. Define E and O if two parity sums are involved.
Calculate P(1) and P(−1) directly from the original polynomial form. Use E = [P(1) + P(−1)]/2 and O = [P(1) − P(−1)]/2.
Check signs when the binomial contains a negative x coefficient, and check the exponent condition for special identities such as E = O = 2^(n−1).
If an unknown parameter is involved, form equations from the requested sums and solve all admissible cases. Verify the final values in the original polynomial rather than trusting a single transformed equation.
Practice questions with checked answers
Question one: find the even and odd coefficient sums of (1 + x)^5. Answer: 16 and 16.
Question two: find the even and odd coefficient sums of (2 + x)^4. Answer: 41 and 40.
Question three: find the even and odd coefficient sums of (1 + 2x)^5. Answer: 121 and 122.
Question four: find the signed even and odd coefficient sums of (2 − x)^4. Answer: 41 and −40.
Question five: the even-power coefficient sum of (1 + x)^n is 32 for a positive integer n. Answer: n = 6.
Question six: find the even and odd coefficient sums of (1 + x²)^5. Answer: 32 and 0.
Question seven: for (1 + x)^4, find the even-indexed and odd-indexed weighted coefficient sums ΣrC(4,r). Answer: 16 and 16.
For every task, write P(1) and P(−1) before considering a full expansion.
Common errors and targeted repair
One student excludes the constant term from the even-power group, forgetting that zero is even. Another treats a negative coefficient’s magnitude as though it were positive.
A third copies E = O for a general binomial (a + bx)^n, even though the two evaluations do not give P(−1) = 0. A fourth tries x = 0, which isolates only the constant.
These are distinct misconceptions. A tutor should ask the learner to state what evaluating at 1, −1 and 0 each does to powers, then test the idea with a small polynomial before moving to a larger binomial.
An independent variation with a negative x coefficient or an x² term is a stronger transfer check than another identical (1 + x)^n example.
Syllabus alignment and further learning
Binomial expansions, coefficients and simple polynomial identities are connected mathematical ideas. Even–odd coefficient separation and derivative-weighted sums may be used as enrichment depending on the school’s assessed topic sequence.
Singapore families can refer to SEAB’s relevant 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2 syllabus listings.
Related guides include How to Recover a Binomial Exponent from Consecutive Coefficients, How to Find a Missing Coefficient in Combined Binomial Expansions and How to Find the Constant Term When a Binomial Contains Reciprocal Powers. More learning resources are available in the Additional Mathematics archive.
Parents may enquire about eduKate Singapore’s maximum three-student Mathematics tutorials near Sixth Avenue MRT. The goal is to recognise when two carefully selected substitutions reveal a coefficient pattern without brute-force expansion. Properly taught kids shine a bright light into the future.