This binomial theorem calculator expands (a + b)ⁿ completely — every term with its binomial coefficient — and can also extract any single term by its position. It handles numeric or simple algebraic first terms, shows the Pascal row it used, and verifies itself the way mathematicians do: by plugging in a = b = 1 and checking that the coefficients sum to 2ⁿ.

Full expansion –
Requested term –
Sanity check (x = 1) –

General term (k starting at 0): T = C(n,k) · aⁿ⁻ᵏ · bᵏ · xⁿ⁻ᵏ. Coefficients C(n,k) are the combinations from Pascal’s triangle — the same numbers our Combination Calculator computes.

The Pattern Hiding Inside (a + b)ⁿ

Multiply out (a + b)² and you get a² + 2ab + b²; cube it and 1, 3, 3, 1 appears; the fourth power carries 1, 4, 6, 4, 1. The binomial theorem is the statement that this is not a coincidence but a counting law: the term aⁿ⁻ᵏbᵏ in (a + b)ⁿ carries the coefficient C(n,k) — the number of ways to choose which k of the n brackets contribute a b — and those coefficients are exactly the rows of Pascal's triangle, each the sum of the two above it. Newton later generalized the whole structure to fractional and negative exponents (where the expansion never terminates), one of the founding results of calculus; the finite case here is the version that runs algebra classes, probability and half of combinatorics. The term-extraction mode answers the second-most-common question about expansions (“what is the coefficient of x³ in (2x − 3)⁷?”) without expanding anything: set k so the x-power matches, evaluate C(n,k)aⁿ⁻ᵏbᵏ, done.

The sanity-check line deserves its moment: setting x = 1 and a = b = 1 in (1 + 1)ⁿ proves the coefficients of any row sum to 2ⁿ — and every row-sum on this page is audited against (a + b)ⁿ directly. Expansions are uniquely error-prone by hand (a dropped sign in term six propagates invisibly), so the calculator shows its audit trail rather than asking for trust.

Worked Example (Verified)

  • (x + 2)⁵ (defaults): coefficients 1, 5, 10, 10, 5, 1 dressed with powers of 2 → x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32. Term 4 (the x² term) = C(5,3)·1²·2³ = 80x² ✔ — the extraction mode's default answer. At x = 1: sum 243 = 3⁵ ✔.
  • (2x − 3)³: set a = 2, b = −3 → 8x³ − 36x² + 54x − 27. Notice the alternating signs ride entirely on b's negativity — odd powers of b flip sign, even powers don't; the theorem mechanizes what students most often fumble.
  • The probability bridge: expand (p + q)⁵ for a fair coin's five flips and the coefficients 1-5-10-10-5-1 ÷ 32 ARE the binomial distribution — five heads 1/32, exactly three heads 10/32. The theorem and the distribution are the same object, which is why “binomial” appears in both names.

Binomial Discipline

  • Term numbering bites: the “fourth term” has k = 3 (k counts b's power, starting from 0). Off-by-one is the most common extraction error in existence; the calculator labels both k and the term number explicitly.
  • Signs belong to b: write (a − b)ⁿ as (a + (−b))ⁿ before starting, always; half of all expansion errors are buried minus signs.
  • The expansion has n + 1 terms — if your answer has n, one was dropped; if coefficients don't sum to (a+b)ⁿ at x = 1, find the error before proceeding.
  • Coefficients are symmetric (C(n,k) = C(n,n−k)) — the expansion reads the same backwards with a and b swapped; use it as a free second check.
  • For (x + 1/x)-style or two-variable expansions, reduce to this form by treating one compound piece as “a” — the theorem doesn't care what the terms contain, only that there are two of them.

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Frequently Asked Questions

What is the binomial theorem formula?

(a + b)ⁿ = Σₖ₌₀ⁿ C(n,k) aⁿ⁻ᵏ bᵏ, where C(n,k) = n!/(k!(n−k)!) is the binomial coefficient — row n of Pascal's triangle. The k-th term (starting k = 0) is C(n,k) aⁿ⁻ᵏ bᵏ. This calculator expands the polynomial form (ax + b)ⁿ fully and can extract any single term, with the coefficient row and an x = 1 audit of the arithmetic.

How do you find a specific term in a binomial expansion?

Use the general term Tₖ₊₁ = C(n,k) aⁿ⁻ᵏ bᵏ with k chosen so the variable's exponent matches your target. For the x² term of (x + 2)⁵: the x-power is n − k = 2, so k = 3, giving C(5,3)·1²·2³·x² = 80x². Remember term numbers are k + 1 (the first term has k = 0) — the off-by-one trap this calculator's term selector is designed to defuse.

How do binomial coefficients relate to Pascal's triangle?

They are Pascal's triangle: row n lists C(n,0)…C(n,n), each interior entry the sum of the two above it (the identity C(n,k) = C(n−1,k−1) + C(n−1,k)). Row 5 is 1, 5, 10, 10, 5, 1 — the bare coefficients of (a + b)⁵ before a's and b's own powers dress them. The triangle also encodes the row-sum 2ⁿ and the symmetry C(n,k) = C(n,n−k), both of which this page uses as built-in error checks.

Why do the coefficients sum to 2ⁿ?

Set a = b = 1: the theorem reads (1 + 1)ⁿ = Σ C(n,k) — the left side is 2ⁿ by arithmetic, so the coefficients must sum to it. Combinatorially it's just as direct: C(n,k) counts the subsets of size k of an n-set, and summing over all sizes counts all subsets, of which there are 2ⁿ. The calculator audits every expansion with this identity at x = 1 (generalized to arbitrary a, b).

How do you expand (a − b)ⁿ?

Rewrite it as (a + (−b))ⁿ and expand normally: the signs then alternate automatically because odd powers of (−b) are negative — (x − 2)³ = x³ − 6x² + 12x − 8. In this calculator, simply enter b as a negative number; the term-by-term output carries the signs explicitly so the alternation is visible rather than memorized.

Where is the binomial theorem used outside algebra class?

Probability (the binomial distribution is the expansion of (p + q)ⁿ read term by term), combinatorics (subset counting), approximation ((1 + x)ⁿ ≈ 1 + nx for small x — the first two terms of the general expansion, underneath half of physics' back-of-envelope work), compound growth derivations, and computer algebra systems, where this exact algorithm expands symbolic powers. Newton's generalization to non-integer exponents extends it into infinite series — the historical doorway to calculus.

What is the middle term of (a + b)ⁿ?

For even n, one middle term at k = n/2 (the largest coefficient of the row — e.g. C(6,3) = 20 in (a+b)⁶). For odd n, two equal-largest central terms at k = (n±1)/2 and (n∓1)/2 (C(5,2) = C(5,3) = 10). Coefficients increase toward the center and mirror away from it — which is why the extraction mode, not the full expansion, is the right tool when only the big central terms matter.

Is this binomial theorem calculator free?

Yes — free, instant, no sign-up. Full expansions to n = 12, single-term extraction, and a built-in coefficient-sum audit on every result.