Positive-integer binomial expansions
Choose valid binomial indices, retain signs and domains, and combine coefficient contributions before solving parameter conditions.
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Each factor contributes one choice
For a positive integer n, (a + b)ⁿ is a product of n identical brackets. Each expanded term chooses either a or b from every bracket. Choosing b from r brackets and a from the remaining n − r gives a term of the form a^(n-r)b^r.
The coefficient counts which r bracket positions supplied b. The notation is
Here r is an integer, n! means n × (n − 1) × … × 1, and 0! = 1. For example, (4 choose 2) = 4!/(2!2!) = 6. The six pairs of b-positions are (1,2), (1,3), (1,4), (2,3), (2,4) and (3,4). Their order within a pair does not create another choice of positions.
Adding the indexed terms gives the binomial theorem:
The formula lists n + 1 indexed terms. Substitute the actual expressions for a and b before collecting powers. Fractional and negative powers need a different expansion and convergence analysis; the finite positive-integer formula here does not apply to them.
For (a + b)⁴, the binomial coefficients are 1, 4, 6, 4 and 1. Thus
For example, the x² term is (4 choose 2) 2²(-x)² = 24x². Its coefficient 24 includes the numerical power 2² as well as the binomial coefficient 6. The odd powers of −x give the negative terms; an even power does not keep a minus sign.
The index r starts at zero
The general term is the (r + 1)th term when listed by increasing choices of b. In (2 − x)⁵, the x³ term uses r = 3:
The coefficient is −40, the term is −40x³, and its position is fourth. These are three different answers to three different questions. A request for the third term would instead use r = 2.
Check your understanding 1 Include both numerical factors
Find the coefficient of x² in (3 + 2x)⁴. State the selected index and term position, and distinguish the coefficient from the whole term.
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Use r = 2, which is the third term. The term is (4 choose 2) 3²(2x)² = 216x², so the coefficient is 216. Taking the binomial coefficient 6 alone omits both 3² and 2²; reporting 216x² would give the term rather than the requested coefficient.
Track the complete variable power
If a and b contain powers of x, choosing r does not necessarily produce xʳ. For (x² + 1/x)⁶, a term has the form
The constant term needs 12 − 3r = 0, giving r = 4. Its coefficient is (6 choose 4) = 15. The original expression still excludes x = 0, even though this particular term is constant. Calling it a constant term does not extend the domain of the whole expression.
The index must be an integer between zero and n. If the exponent equation gives a non-integer or out-of-range r, that power has no term in this indexed expansion. Do not round the index or use an unrelated binomial coefficient.
Check your understanding 2 An exponent equation can have no allowed index
For (x² + 1/x)⁵, find the coefficient of x⁴ and decide whether any x⁵ term occurs. Keep the original domain and justify each selected or rejected index.
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The exponent is 2(5 − r) − r = 10 − 3r. For x⁴, r = 2 is allowed, so its coefficient is (5 choose 2) = 10. For x⁵, the equation gives r = 5/3, which is not an integer. There is no x⁵ term and its coefficient is zero. The full expression requires x ≠ 0 in both conclusions.
A coefficient condition can leave more than one parameter
If the coefficient of x² in (1 + kx)⁵ is 40, then 10k² = 40, so k = 2 or −2. The even power alone does not select the sign. If the x³ coefficient is also stated to be negative, 10k³ < 0 selects k = −2. Its x³ coefficient is then −80.
Use every stated condition and retain all candidates until a condition rejects them. Selecting a positive square root without a reason would discard a valid parameter from the first condition alone.
Verify a short expansion
Check your understanding 3 Use substitutions as checks, rather than as a proof
Expand (1 − 2x)³ and check at x = 0 and x = 1. Explain what those substitutions can and cannot establish about an expansion claimed for every x.
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The expansion is 1-6x + 12x²-8x³. At zero it gives 1; at one it gives −1, matching (−1)³. A failed check disproves the proposed identity, but agreement at two inputs alone does not prove it for every x. The binomial selection argument or complete distribution establishes the identity.
Try it yourself 1
Recover the parameter from a cancelled coefficient
Problem
Let P(x) = (1 − 2x)(1 + ax)⁵, where a is a nonzero real constant. The coefficient of x² in P is zero. Find a and the coefficient of x³. A proposal takes the x³ coefficient from the fifth power alone as (5 choose 3)a³. Decide whether it gives the coefficient in P. Explain whether the zero x² coefficient means that no x² contributions arose during multiplication.
Write your answer, then compare it with the solution
Hints
Hint 1
Inspect every way the outer bracket can produce the requested power
A term of the requested power can come from multiplying by 1 or from multiplying a lower power by −2x. Collect these contributions before imposing the zero-coefficient condition. Keep the given restriction on a when selecting a candidate.
Show full solution
Form the complete x² coefficient
Method
Combine the contributions from both terms of the outer bracket.Reason
Multiplication by −2x raises the inner power by one.Working
Answer
The inner x² coefficient is 10a², while its x coefficient is 5a. Thus the x² coefficient in P is 10a² − 2(5a) = 10a² − 10a. Using only the inner coefficient would miss the second contribution.Use the coefficient condition and parameter restriction
Method
Set the collected coefficient equal to zero.Reason
A coefficient that cancels must be evaluated after multiplication, and a was stated to be nonzero.Working
Answer
10a(a − 1) = 0 gives a = 0 or 1. The nonzero condition excludes 0, so a = 1. The excluded value would satisfy the coefficient equation but would not satisfy the original problem.Find the x³ coefficient in the full product
Method
Use the inner x³ and x² coefficients.Reason
Both outer terms can contribute to x³ too.Working
Answer
The coefficient is 10a³ − 2(10a²). At a = 1 this is 10 − 20 = −10. The corresponding term is −10x³. The proposal gives +10 because it ignores the contribution from −2x.Explain cancellation rather than absence
Method
Check just the relevant beginning of the product at a = 1.Reason
A collected coefficient can be zero even when intermediate contributions are nonzero.Working
Answer
The fifth power begins 1 + 5x + 10x² + 10x³, followed by higher powers. Multiplication by 1 − 2x gives the initial terms 1 + 3x + 0x² − 10x³. The x² contributions +10x² and −10x² cancel; they were not absent from the multiplication.
Quick check
Compare the selected index, complete numerical and variable powers, every outer-factor contribution and the parameter restriction. A different valid expansion method is welcome. These are self-review criteria: distinguish a missing indexed power, a coefficient cancelled by collection, a term and its numerical coefficient.
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