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Binomial Theorem



The Theorem
The General Term
Special Cases
The Multinomial Theorem
Binomial Expansion Notation
Worked Expansions
Related Concepts
Binomial Theorem at a Glance
Binomial Theorem FAQ



Where Counting Meets Algebra


The binomial coefficient was defined to count subsets — the number of kk-element selections from an nn-element set. The same coefficient also organizes the algebraic expansion of (a+b)n(a+b)^n, and the binomial theorem makes that organization explicit.

The connection is not a coincidence. When (a+b)n(a+b)^n is written out as the product of nn copies of (a+b)(a+b), each term in the expansion is built by choosing either aa or bb from each factor and multiplying the choices together. A term of the form an−kbka^{n-k} b^k arises whenever exactly kk of the nn factors contribute a bb, and the number of ways to make that choice is (nk)\binom{n}{k}.

Counting subsets and expanding powers turn out to be the same problem viewed from two sides.



The Theorem


For any non-negative integer nn and any values aa and bb,

Binomial Theorem
(a+b)n=∑k=0n(nk)an−kbk.(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.
Learn more about this formula: Binomial Theorem →


The expansion has exactly n+1n+1 terms, indexed by kk from 00 through nn. The coefficients (n0),(n1),…,(nn)\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n} are precisely the entries of row nn of Pascal's triangle — so the row of the triangle reads as the row of coefficients in the expansion. The ∑\sum summation sign and the rest of the theorem's notation are collected among the combinatorics symbols.

Combinatorial Proof


Write (a+b)n(a+b)^n as the product

(a+b)(a+b)⋯(a+b)⏟n factors.\underbrace{(a+b)(a+b)\cdots(a+b)}_{n \text{ factors}}.


Expanding the product means forming every possible term by choosing either aa or bb from each factor and multiplying the selections. A term equal to an−kbka^{n-k} b^k arises whenever exactly kk of the nn factors contribute a bb, and there are (nk)\binom{n}{k} ways to make that selection. Summing over all kk gives the theorem.

This is a double counting argument in its purest form: the product (a+b)n(a+b)^n, viewed both as an algebraic expression and as an enumeration of choice-sequences, has the same value either way.

Algebraic Proof


An induction on nn also works, using Pascal's rule to advance from row nn to row n+1n+1. Both proofs are standard; the combinatorial one is the more direct explanation of why binomial coefficients appear.
The 8 picks from (a + b)(a + b)(a + b), grouped by the number of b’spick a or b from each factor of (a + b)(a + b)(a + b)8 = 2 × 2 × 2 picks in all, sorted by how many b’s they takeno baaa1 pick = C(3,0)a³+one baabababaa3 picks = C(3,1)3a²b+two b’sabbbabbba3 picks = C(3,2)3ab²+three b’sbbb1 pick = C(3,3)b³(a + b)³ = a³ + 3a²b + 3ab² + b³
Every pick, sorted by its b's

Multiplying out (a + b)(a + b)(a + b) means taking a or b from each of the three factors, 2 × 2 × 2 = 8 picks in all. Sort the picks by how many b’s they take: one pick has none, three have one, three have two, one has three. Each column is a set of positions chosen for the b’s, so its size is C(3, k), and all its picks multiply to the same a³⁻ᵏbᵏ — which is why that binomial coefficient becomes the term’s coefficient.

The coefficients 1, 3, 3, 1 are row 3 of Pascal's triangle; with four factors the same sorting gives 1, 4, 6, 4, 1.

The General Term


The term corresponding to a specific power of bb in the expansion appears often enough to deserve its own formula. The (k+1)(k+1)-th term is

General Term in Binomial Expansion
Tk+1=(nk)an−kbk.T_{k+1} = \binom{n}{k} a^{n-k} b^k.
Learn more about this formula: General Term in Binomial Expansion →


Index Convention


The terms are numbered starting from T1=(n0)anT_1 = \binom{n}{0} a^n — the term with no bb. Under this convention, the term containing bkb^k is Tk+1T_{k+1}, not TkT_k.

Uses


The general term is the tool for problems that ask about one specific term rather than the whole expansion. Three standard problem types use it directly:

• Finding the coefficient of a specific power of one variable
• Identifying the middle term of the expansion
• Finding the term independent of a variable (when one of aa or bb involves that variable in the numerator and the other in the denominator)

The Middle Term


When nn is even, the expansion has n+1n+1 terms with a unique middle term

Middle Term in Binomial Expansion
Tn/2+1=(nn/2)an/2bn/2,n even.T_{n/2 + 1} = \binom{n}{n/2} a^{n/2} b^{n/2}, \quad n \text{ even}.
Learn more about this formula: Middle Term in Binomial Expansion →


Its coefficient (nn/2)\binom{n}{n/2} is the maximum binomial coefficient in row nn — the entries grow from the edges toward the middle of each row of Pascal's triangle.

Special Cases


Specific substitutions for aa and bb convert the theorem into useful identities.

Setting a=1,b=1a = 1, b = 1


(1+1)n=2n=∑k=0n(nk).(1+1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k}.


This is the algebraic counterpart of the row-sum identity from Pascal's triangle: the entries of row nn sum to 2n2^n, which is also the number of subsets of an nn-set.

Setting a=1,b=−1a = 1, b = -1


(1−1)n=0=∑k=0n(−1)k(nk)for n≥1.(1-1)^n = 0 = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \quad \text{for } n \ge 1.


The alternating row sum is zero, which means the number of even-sized subsets of an nn-set equals the number of odd-sized subsets.

Setting a=1,b=xa = 1, b = x


(1+x)n=∑k=0n(nk)xk.(1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k.


This is the form most commonly encountered in algebra. The right side is the generating function for the binomial coefficients of row nn — a polynomial whose coefficients encode the entire row.

Derived Identities


Operations on the polynomial form produce further identities. Differentiating (1+x)n(1+x)^n with respect to xx and substituting x=1x = 1 gives

Weighted Binomial Sum
∑k=1nk(nk)=n⋅2n−1.\sum_{k=1}^{n} k \binom{n}{k} = n \cdot 2^{n-1}.
Learn more about this formula: Weighted Binomial Sum →


Multiplying by xx and differentiating again, or integrating and substituting, produces a family of related sums. These substitutions are the standard route for converting algebraic facts about (a+b)n(a+b)^n into combinatorial identities about binomial coefficients.
Substitution Identity produced Combinatorial meaning
a = 1, b = 1 Σk=0n C(n, k) = 2n row sum = total number of subsets of an n-set
a = 1, b = −1 Σk=0n (−1)k C(n, k) = 0 for n ≥ 1 even-sized subsets are as numerous as odd-sized ones
a = 1, b = x (1 + x)n = Σk=0n C(n, k) xk generating function for row n of Pascal's triangle
Differentiate, then x = 1 Σk=1n k · C(n, k) = n · 2n−1 weighted row sum; combinatorial proofs by double counting also work
n = 0n = 1n = 2n = 3n = 4sum = 16 = 2⁴n = 5n = 61111211331146C(4, 2)41151010511615201561ROW 4 = COEFFICIENTS OF (a + b)⁴(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
Row 4 sums to 16 = 2⁴

The whole of row 4 is highlighted and its entries total 16, which is the substitution a = b = 1 carried out in the array instead of in the algebra. Reading the same row as a count of subsets explains why the total is a power of two: each of the 4 items is independently either in or out. Highlight any row and watch its total double on the Pascal’s triangle visualizer.

The other substitutions work the same way: each one turns a free choice of a and b into a statement about the coefficients themselves.

The Multinomial Theorem


The binomial theorem expands a sum of two terms raised to a power. When the base has more than two terms, the generalization is the multinomial theorem.

Statement


For a non-negative integer nn and any values x1,x2,…,xrx_1, x_2, \ldots, x_r,

Multinomial Theorem
(x1+x2+⋯+xr)n=∑k1+k2+⋯+kr=nki≥0(nk1,k2,…,kr)x1k1x2k2⋯xrkr.(x_1 + x_2 + \cdots + x_r)^n = \sum_{\substack{k_1 + k_2 + \cdots + k_r = n \\ k_i \ge 0}} \binom{n}{k_1, k_2, \ldots, k_r} x_1^{k_1} x_2^{k_2} \cdots x_r^{k_r}.
Learn more about this formula: Multinomial Theorem →


where the sum runs over all tuples (k1,k2,…,kr)(k_1, k_2, \ldots, k_r) of non-negative integers with k1+k2+⋯+kr=nk_1 + k_2 + \cdots + k_r = n.

The Coefficient


The coefficient is the multinomial coefficient

(nk1,k2,…,kr)=n!k1! k2! ⋯ kr!,\binom{n}{k_1, k_2, \ldots, k_r} = \frac{n!}{k_1! \, k_2! \, \cdots \, k_r!},


which counts the number of ways to partition nn items into rr labeled groups of the specified sizes. The binomial theorem is the case r=2r = 2: the coefficient (nk, n−k)\binom{n}{k, \, n-k} collapses to the ordinary binomial coefficient (nk)\binom{n}{k}.

Number of Terms


The number of distinct terms in the multinomial expansion equals the number of tuples of non-negative integers summing to nn, which is a weak composition count:

(n+r−1r−1).\binom{n+r-1}{r-1}.


For the binomial case r=2r = 2 this gives (n+11)=n+1\binom{n+1}{1} = n+1, matching the count from the binomial theorem.
4!/(2! · 1! · 1!) = 12Complete · 12 / 12SOURCE (n = 4)1234BUILD BOXES (2+1+1)BOX A (2)BOX B (1)BOX C (1)COMPLETED12 / 12→ BoxA123412431324134214231432→ BoxB231424133412→ BoxC234124313421
Four items into boxes of 2, 1 and 1

The coefficient of x₁²x₂x₃ in (x₁ + x₂ + x₃)⁴ counts the ways to hand the 4 factors to three labelled groups: two factors give x₁, one gives x₂, one gives x₃. Filling Box A with 2 of the 4 items, then Box B and Box C with 1 each, gives 4!/(2!·1!·1!) = 12 partitions, so the term is 12x₁²x₂x₃. Try other group sizes on the partition into groups visualizer.

With only two boxes, of sizes k and n - k, the same count is the binomial coefficient: the binomial theorem is the two-group case.

Binomial Expansion Notation

Notation

Binomial Expansion Notation

The theorem's machinery in three marks: the sigma that folds an expansion into one expression, the term labels with their built-in shift, and the sum whose subscript is an equation.
(nk)\binom{n}{k} and the multinomial coefficient are owned by binomial coefficient notation, n!n! by permutation notation; the alternating (−1)k(-1)^k of Special Cases above is the device solution sets formalize.
∑k=0n\sum_{k=0}^{n}
the sum as k runs from zero to n
The expansion compressor: capital sigma with its counter declared below and its stopping point above — n+1n + 1 terms folded into one expression in The Theorem above. The letter is an operator, not a quantity.
CasesThe counter kk is a bound (dummy) variable — the sum's value never mentions it, the same disappearing act as the dummy variable of integration; renaming kk to jj changes nothing.
Do not confuseA product of sigma and its neighbors. Nothing multiplies — everything to the right of the sigma sits inside its scope; how far the scope reaches is convention, and brackets settle any dispute.
Tk+1T_{k+1}
the k-plus-first term
Term labels with a built-in shift: numbering starts at T1T_1 — the term with no bb — so the term containing bkb^k is Tk+1T_{k+1}, never TkT_k; the Index Convention of The General Term above.
CasesThe middle term inherits the shift — even nn puts it at Tn/2+1T_{n/2+1}; every "find the 7th term" problem is really a k=6k = 6 computation, and writing that translation down first prevents the classic slip.
Do not confuseZero-based counting. The exponent kk runs from 00, the label TT from 11 — two counters one step apart riding the same term; conflating them is the standard error of expansion problems.
∑k1+⋯+kr=n\sum_{k_1 + \cdots + k_r = n}
the sum over all splittings of n
Summation with an equation for a subscript: the index set is every non-negative solution of k1+⋯+kr=nk_1 + \cdots + k_r = n — one term per way of splitting nn, as The Multinomial Theorem above deploys it.
CasesThe constraint replaces bounds entirely — no corners, just a membership condition under the sigma, with extra conditions stacked beneath (ki≥0k_i \geq 0): a miniature set-builder doing index duty.
Do not confuseA single running index. No one counter exists here — the sum ranges over tuples; expanding it as if one kk walks from somewhere to nn produces phantom terms and misses most of the real ones.

Worked Expansions


Four worked examples covering the standard problem types.

Full Expansion: (x+2)4(x+2)^4


Substitute a=xa = x and b=2b = 2. The coefficients are row 4 of Pascal's triangle: 1,4,6,4,11, 4, 6, 4, 1. The expansion is

(x+2)4=x4+4x3(2)+6x2(4)+4x(8)+16=x4+8x3+24x2+32x+16.(x+2)^4 = x^4 + 4x^3(2) + 6x^2(4) + 4x(8) + 16 = x^4 + 8x^3 + 24x^2 + 32x + 16.


Expansion with Signs: (2a−3b)3(2a - 3b)^3


Substitute the first base as 2a2a and the second as −3b-3b. Row 3 coefficients are 1,3,3,11, 3, 3, 1:

(2a−3b)3=(2a)3+3(2a)2(−3b)+3(2a)(−3b)2+(−3b)3.(2a - 3b)^3 = (2a)^3 + 3(2a)^2(-3b) + 3(2a)(-3b)^2 + (-3b)^3.


Simplifying each term:

(2a−3b)3=8a3−36a2b+54ab2−27b3.(2a - 3b)^3 = 8a^3 - 36a^2 b + 54a b^2 - 27 b^3.


The alternating signs come directly from the powers of −3b-3b.

Single Coefficient: x5x^5 in (1+x)8(1+x)^8


The general term of (1+x)8(1+x)^8 is Tk+1=(8k)xkT_{k+1} = \binom{8}{k} x^k. The coefficient of x5x^5 comes from k=5k = 5:

(85)=56.\binom{8}{5} = 56.


No need to expand the full polynomial.

Term Independent of xx in (x+1x)10\left(x + \frac{1}{x}\right)^{10}


The general term is

Tk+1=(10k)x10−k(1x)k=(10k)x10−2k.T_{k+1} = \binom{10}{k} x^{10-k} \left(\frac{1}{x}\right)^k = \binom{10}{k} x^{10 - 2k}.


The term is independent of xx when 10−2k=010 - 2k = 0, i.e. k=5k = 5. The constant term is

(105)=252.\binom{10}{5} = 252.


Multinomial Expansion: (a+b+c)3(a+b+c)^3


The tuples (k1,k2,k3)(k_1, k_2, k_3) of non-negative integers summing to 3 are: (3,0,0)(3,0,0), (0,3,0)(0,3,0), (0,0,3)(0,0,3), (2,1,0)(2,1,0), (2,0,1)(2,0,1), (1,2,0)(1,2,0), (0,2,1)(0,2,1), (1,0,2)(1,0,2), (0,1,2)(0,1,2), (1,1,1)(1,1,1). The corresponding coefficients are 1,1,1,3,3,3,3,3,3,61, 1, 1, 3, 3, 3, 3, 3, 3, 6:

(a+b+c)3=a3+b3+c3+3a2b+3a2c+3ab2+3b2c+3ac2+3bc2+6abc.(a+b+c)^3 = a^3 + b^3 + c^3 + 3a^2 b + 3a^2 c + 3a b^2 + 3 b^2 c + 3a c^2 + 3 b c^2 + 6abc.


The number of terms, (3+3−13−1)=(52)=10\binom{3+3-1}{3-1} = \binom{5}{2} = 10, matches the count of distinct monomials above.
Problem type Setup Key step Answer
Full expansion (x + 2)4; row 4 = 1, 4, 6, 4, 1 substitute b = 2 into each term x4 + 8x3 + 24x2 + 32x + 16
Expansion with signs (2a − 3b)3; row 3 = 1, 3, 3, 1 track the sign of (−3b)k in each term 8a3 − 36a2b + 54ab2 − 27b3
One specific coefficient coefficient of x5 in (1 + x)8 general term gives k = 5; compute C(8, 5) 56
Term independent of x (x + 1⁄x)10 solve 10 − 2k = 0 → k = 5 C(10, 5) = 252
Multinomial expansion (a + b + c)3 enumerate all (k₁, k₂, k₃) with k₁+k₂+k₃ = 3 10 distinct monomials including 6abc

Related Concepts


• Polynomials — polynomial identities, factoring, and the structure of expansions in algebra rest on the binomial theorem and its generalizations.

• Binomial Distribution — the binomial distribution in probability is constructed directly from the binomial theorem; the probabilities of all possible outcome counts sum to 11 by the substitution a=pa = p, b=1−pb = 1-p.

Binomial Theorem at a Glance


The page covered the theorem statement, the general-term formula, useful substitutions that yield combinatorial identities, the multinomial generalization, and the standard problem types. The table below collects the structural facts about (a+b)n in one reference card.
Concept Statement Example
The theorem (a + b)n = Σk=0n C(n, k) · an−k · bk (x + y)3 = x3 + 3x2y + 3xy2 + y3
Number of terms exactly n + 1, one for each k from 0 to n (a + b)5 has 6 terms
General term Tk+1 = C(n, k) · an−k · bk; indexing starts at T1 = an T3 in (a + b)4 is C(4, 2) a2b2 = 6a2b2
Coefficients = Pascal row the n+1 coefficients are row n of Pascal's triangle (a + b)4: 1, 4, 6, 4, 1
Coefficient symmetry C(n, k) = C(n, n − k) — each row reads the same forwards and backwards 1, 4, 6, 4, 1 is palindromic
Middle term (n even) one middle term Tn/2 + 1 with the largest coefficient C(n, n/2) (a + b)6: T4 = 20 a3b3
Middle terms (n odd) two middle terms T(n+1)/2 and T(n+3)/2 with equal coefficients (a + b)5: T3 and T4, both coefficient 10
Multinomial generalization (x₁ + … + xr)n = Σ C(n; k₁, …, kr) · x₁k₁…xrkr with k₁ + … + kr = n (a + b + c)3: 10 distinct monomials
Multinomial term count distinct monomials = C(n + r − 1, r − 1) — a weak-composition count (a+b+c)3: C(5, 2) = 10
Where it appears polynomial expansions, binomial distribution (substitute a = p, b = 1−p), Newton's generalized binomial series for non-integer n (p + (1−p))n = 1 sums all binomial probabilities

Binomial Theorem FAQ

How many terms does the expansion of (a + b) to the n have?

+
One more than the exponent. The powers of the first term run from n down to zero, giving n plus one values in total, and each contributes exactly one term. Squaring gives three terms, cubing gives four. Terms only merge if the two parts happen to be equal, which the general form does not assume.Read more →

How do you find one specific term without expanding everything?

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Use the general term formula, which gives any term directly from its position. Be careful with indexing: the term numbered r plus one carries the exponent r, so asking for the fifth term means setting r to four. Off-by-one errors here are the most common mistake in the whole topic.Read more →

How do you find the middle term of a binomial expansion?

+
When the exponent is even there is a single middle term, sitting at position n over two plus one. When the exponent is odd the count of terms is even and there are two middle terms instead. Checking the parity of the exponent before hunting for the position avoids looking for a term that does not exist.Read more →

What happens to the signs when you expand (a − b) to the n?

+
They alternate, starting positive. Treating the expression as a plus negative b means each term picks up a factor of negative one raised to the term's index, so odd-numbered powers of b come out negative. The coefficients themselves are unchanged; only the signs differ from the sum case.Read more →