What is the binomial theorem?

The binomial theorem states that for any non-negative integer n, the expansion of (a+b)^n is the sum from k=0 to n of C(n,k)·a^(n-k)·b^k. The coefficients are the binomial coefficients, which are exactly the entries of row n of Pascal's triangle.

What is the general term in a binomial expansion?

The general term, also called the (k+1)-th term, is T_(k+1) = C(n,k)·a^(n-k)·b^k. Indexing starts at T_1 = a^n. The general term is used to find a specific coefficient, the middle term, or the term independent of a variable without expanding the full polynomial.

How many terms does (a+b)^n have?

The expansion of (a+b)^n has exactly n+1 terms, one for each value of k from 0 to n. When n is even, there is a unique middle term T_(n/2+1) with the maximum coefficient C(n, n/2).

What is the multinomial theorem?

The multinomial theorem generalizes the binomial theorem to sums of more than two terms. It expands (x_1 + x_2 + ... + x_r)^n as a sum over all tuples (k_1, ..., k_r) of non-negative integers summing to n, with multinomial coefficients n!/(k_1!·k_2!·...·k_r!).

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

When n is even, the middle term of (a+b)^n is the (n/2 + 1)-th term: T_(n/2+1) = C(n, n/2)·a^(n/2)·b^(n/2). Its coefficient C(n, n/2) is the largest in row n of Pascal's triangle. When n is odd, there are two middle terms with equal coefficients.

Binomial Theorem

The Theorem

The General Term

Special Cases

The Multinomial Theorem

Worked Expansions

Related Concepts

Binomial Theorem at a Glance

Where Counting Meets Algebra

The binomial coefficient was defined to count subsets — the number of

k

-element selections from an

n

-element set. The same coefficient also organizes the algebraic expansion of

(a+b)^n

, and the binomial theorem makes that organization explicit.

The connection is not a coincidence. When

(a+b)^n

is written out as the product of

n

copies of

(a+b)

, each term in the expansion is built by choosing either

a

b

from each factor and multiplying the choices together. A term of the form

a^{n-k} b^k

arises whenever exactly

k

of the

n

factors contribute a

b

, and the number of ways to make that choice is

\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

n

and any values

a

and

b

Binomial Theorem

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.

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The expansion has exactly

n+1

terms, indexed by

k

from

0

through

n

. The coefficients

\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}

are precisely the entries of row

n

of Pascal's triangle — so the row of the triangle reads as the row of coefficients in the expansion.

Combinatorial Proof

Write

(a+b)^n

as the product

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

Expanding the product means forming every possible term by choosing either

a

b

from each factor and multiplying the selections. A term equal to

a^{n-k} b^k

arises whenever exactly

k

of the

n

factors contribute a

b

, and there are

\binom{n}{k}

ways to make that selection. Summing over all

k

gives the theorem.

This is a double counting argument in its purest form: the product

(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

n

also works, using Pascal's rule to advance from row

n

to row

n+1

. Both proofs are standard; the combinatorial one is the more direct explanation of why binomial coefficients appear.

The General Term

The term corresponding to a specific power of

b

in the expansion appears often enough to deserve its own formula. The

(k+1)

-th term is

General Term in Binomial Expansion

T_{k+1} = \binom{n}{k} a^{n-k} b^k.

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Index Convention

The terms are numbered starting from

T_1 = \binom{n}{0} a^n

— the term with no

b

. Under this convention, the term containing

b^k

T_{k+1}

, not

T_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

a

b

involves that variable in the numerator and the other in the denominator)

The Middle Term

When

n

is even, the expansion has

n+1

terms with a unique middle term

Middle Term in Binomial Expansion

T_{n/2 + 1} = \binom{n}{n/2} a^{n/2} b^{n/2}, \quad n \text{ even}.

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Its coefficient

\binom{n}{n/2}

is the maximum binomial coefficient in row

n

— the entries grow from the edges toward the middle of each row of Pascal's triangle.

Special Cases

Specific substitutions for

a

and

b

convert the theorem into useful identities.

Setting $a = 1, b = 1$

(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

n

sum to

2^n

, which is also the number of subsets of an

n

-set.

Setting $a = 1, b = -1$

(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

n

-set equals the number of odd-sized subsets.

Setting $a = 1, b = x$

(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

n

— a polynomial whose coefficients encode the entire row.

Derived Identities

Operations on the polynomial form produce further identities. Differentiating

(1+x)^n

with respect to

x

and substituting

x = 1

gives

Weighted Binomial Sum

\sum_{k=1}^{n} k \binom{n}{k} = n \cdot 2^{n-1}.

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Multiplying by

x

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

into combinatorial identities about binomial coefficients.

Substitution	Identity produced	Combinatorial meaning
a = 1, b = 1	Σ_k=0ⁿ C(n, k) = 2ⁿ	row sum = total number of subsets of an n-set
a = 1, b = −1	Σ_k=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 = Σ_k=0ⁿ C(n, k) x^k	generating function for row n of Pascal's triangle
Differentiate, then x = 1	Σ_k=1ⁿ k · C(n, k) = n · 2ⁿ⁻¹	weighted row sum; combinatorial proofs by double counting also work

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

n

and any values

x_1, x_2, \ldots, x_r

Multinomial Theorem

(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}.

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where the sum runs over all tuples

(k_1, k_2, \ldots, k_r)

of non-negative integers with

k_1 + k_2 + \cdots + k_r = n

The Coefficient

The coefficient is the multinomial coefficient

\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

n

items into

r

labeled groups of the specified sizes. The binomial theorem is the case

r = 2

: the coefficient

\binom{n}{k, \, n-k}

collapses to the ordinary binomial coefficient

\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

n

, which is a weak composition count:

\binom{n+r-1}{r-1}.

For the binomial case

r = 2

this gives

\binom{n+1}{1} = n+1

, matching the count from the binomial theorem.

Worked Expansions

Four worked examples covering the standard problem types.

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

Substitute

a = x

and

b = 2

. The coefficients are row 4 of Pascal's triangle:

1, 4, 6, 4, 1

. The expansion is

(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: $(2a - 3b)^3$

Substitute the first base as

2a

and the second as

-3b

. Row 3 coefficients are

1, 3, 3, 1

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

Simplifying each term:

(2a - 3b)^3 = 8a^3 - 36a^2 b + 54a b^2 - 27 b^3.

The alternating signs come directly from the powers of

-3b

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

The general term of

(1+x)^8

T_{k+1} = \binom{8}{k} x^k

. The coefficient of

x^5

comes from

k = 5

\binom{8}{5} = 56.

No need to expand the full polynomial.

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

The general term is

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

x

when

10 - 2k = 0

, i.e.

k = 5

. The constant term is

\binom{10}{5} = 252.

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

The tuples

(k_1, k_2, k_3)

of non-negative integers summing to 3 are:

(3,0,0)

(0,3,0)

(0,0,3)

(2,1,0)

(2,0,1)

(1,2,0)

(0,2,1)

(1,0,2)

(0,1,2)

(1,1,1)

. The corresponding coefficients are

1, 1, 1, 3, 3, 3, 3, 3, 3, 6

(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,

\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

1

by the substitution

a = p

b = 1-p