In short

The binomial theorem states that for any positive integer n, (a + b)^n = \displaystyle\sum_{r=0}^{n} \binom{n}{r}\, a^{n-r}\, b^r. Each term has the form \binom{n}{r}\, a^{n-r}\, b^r, where \binom{n}{r} is the binomial coefficient from combinations. The coefficients form the rows of Pascal's triangle, and the theorem turns the tedious task of multiplying n brackets into a single formula.

Suppose you are computing compound interest. A principal of P rupees grows to P(1 + r)^n after n years at rate r. For n = 2 or n = 3, you can expand (1 + r)^n by hand using algebraic identities. But what about n = 10? Or n = 20? Multiplying twenty brackets of (1 + r) is not a task anyone wants to do manually.

The binomial theorem gives you a direct formula for the expansion of (a + b)^n for any positive integer n. No repeated multiplication, no guesswork about coefficients. Each coefficient turns out to be a combination \binom{n}{r} — the same numbers that count selections of r objects from n.

Why are combinations hiding inside an algebra problem?

Start with something small. Expand (a + b)^3 by writing it as three factors:

(a + b)^3 = (a + b)(a + b)(a + b)

When you multiply these out, every term in the final expansion comes from choosing either a or b from each of the three brackets and multiplying those choices together. For example:

Each term has the form a^{3-r}b^r, where r is the number of brackets from which you chose b. The coefficient of a^{3-r}b^r is the number of ways to choose which r of the 3 brackets contribute a b — that is \binom{3}{r}.

Expanding (a+b) cubed by choosing a or b from each bracketThree brackets labelled (a+b) are shown at the top. Below, four rows show each possible product: choosing b from 0, 1, 2, or 3 brackets. The number of ways to make each choice is C(3,0)=1, C(3,1)=3, C(3,2)=3, C(3,3)=1, giving coefficients 1, 3, 3, 1. (a + b) × (a + b) × (a + b) bracket 1 bracket 2 bracket 3 r = 0: a a a → a³ ×1 r = 1: b a a → a²b ×3 r = 2: b b a → ab² ×3 r = 3: b b b → b³ ×1 (a+b)³ = 1·a³ + 3·a²b + 3·ab² + 1·b³ coefficients: C(3,0), C(3,1), C(3,2), C(3,3) = 1, 3, 3, 1
Each term of $(a+b)^3$ comes from choosing $a$ or $b$ from each of the three brackets. The coefficient of $a^{3-r}b^r$ counts the $\binom{3}{r}$ ways to choose which $r$ brackets contribute $b$. The row shown — one representative choice per row — is just one of the $\binom{3}{r}$ ways.

This pattern works for every positive integer n. Each term of (a+b)^n comes from a binary choice across n brackets, and the coefficient counts those choices.

The theorem

Binomial Theorem (positive integer exponent)

For every positive integer n and any a, b:

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r}\, a^{n-r}\, b^r

Written out:

(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n

There are n + 1 terms. In each term, the exponents of a and b add up to n.

The expansion of (a + b)^4, for example, has 5 terms:

(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

with coefficients \binom{4}{0}, \binom{4}{1}, \binom{4}{2}, \binom{4}{3}, \binom{4}{4} = 1, 4, 6, 4, 1.

Expansion of (a+b) to the fourth power with binomial coefficientsFive terms are listed: C(4,0) a to the 4, C(4,1) a cubed b, C(4,2) a squared b squared, C(4,3) a b cubed, and C(4,4) b to the 4. Below each, the numerical coefficient is shown: 1, 4, 6, 4, 1. (a + b)⁴ — five terms C(4,0)a⁴ C(4,1)a³b C(4,2)a²b² C(4,3)ab³ C(4,4)b⁴ 1 4 6 4 1 a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ exponents in each term add to 4
The five terms of $(a+b)^4$. The coefficient pattern $1, 4, 6, 4, 1$ is the $n = 4$ row of Pascal's triangle. Each coefficient $\binom{4}{r}$ counts the ways to pick $r$ of the four brackets to contribute $b$.

Proof by induction

The combinatorial argument above (choosing a or b from each bracket) is convincing, but a formal proof by mathematical induction makes it watertight.

Base case. For n = 1: (a + b)^1 = a + b = \binom{1}{0}a + \binom{1}{1}b. The formula holds.

Inductive step. Assume the theorem holds for some positive integer k:

(a + b)^k = \sum_{r=0}^{k} \binom{k}{r}\, a^{k-r}\, b^r

Multiply both sides by (a + b):

(a + b)^{k+1} = \sum_{r=0}^{k} \binom{k}{r}\, a^{k-r}\, b^r \cdot (a + b)
= \sum_{r=0}^{k} \binom{k}{r}\, a^{k+1-r}\, b^r + \sum_{r=0}^{k} \binom{k}{r}\, a^{k-r}\, b^{r+1}

In the second sum, substitute s = r + 1 (so r = s - 1, and as r runs from 0 to k, s runs from 1 to k + 1):

= \sum_{r=0}^{k} \binom{k}{r}\, a^{k+1-r}\, b^r + \sum_{s=1}^{k+1} \binom{k}{s-1}\, a^{k+1-s}\, b^{s}

The r = 0 term of the first sum gives \binom{k}{0}a^{k+1} = a^{k+1}. The s = k+1 term of the second sum gives \binom{k}{k}b^{k+1} = b^{k+1}. For 1 \leq r \leq k, the two sums contribute terms with the same powers a^{k+1-r}b^r, and their coefficients add:

\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}

This is Pascal's identity — proven in properties of combinations. Collecting all terms:

(a+b)^{k+1} = \sum_{r=0}^{k+1} \binom{k+1}{r}\, a^{k+1-r}\, b^r

The formula holds for k + 1. By induction, it holds for all positive integers n.

The general term

General term of the binomial expansion

The (r+1)-th term of the expansion of (a + b)^n is:

T_{r+1} = \binom{n}{r}\, a^{n-r}\, b^r \qquad (r = 0, 1, 2, \dots, n)

The index r starts at 0, so T_1 = \binom{n}{0}a^n, T_2 = \binom{n}{1}a^{n-1}b, and so on up to T_{n+1} = \binom{n}{n}b^n.

The general term formula is the workhorse. To find any specific term, plug in n, a, b, and r.

Finding a specific term. The 5th term of (2x + 3)^7:

Here n = 7, a = 2x, b = 3, and the 5th term has r = 4 (since the (r+1)-th term corresponds to r):

T_5 = \binom{7}{4}(2x)^{7-4}(3)^4 = 35 \cdot 8x^3 \cdot 81 = 22{,}680\,x^3
Structure of the general term T sub r+1A diagram showing the three components of the general term: the binomial coefficient C(n,r), the power of a which is a to the n minus r, and the power of b which is b to the r. All three multiply together to give the (r+1)-th term. Anatomy of the general term T(r+1) C(n, r) coefficient × a^(n−r) power of a × b^r power of b exponents (n − r) + r = n always
Every term in the expansion is built from three pieces: the binomial coefficient $\binom{n}{r}$, the descending power of $a$, and the ascending power of $b$. The two exponents always sum to $n$.

Pascal's triangle and the binomial coefficients

The coefficients \binom{n}{r} arrange themselves into Pascal's triangle — a triangular array where each entry is the sum of the two entries directly above it:

\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}
Pascal's triangle rows 0 through 6 with binomial coefficientsA triangular array of numbers. Row 0 has 1. Row 1 has 1 1. Row 2 has 1 2 1. Row 3 has 1 3 3 1. Row 4 has 1 4 6 4 1. Row 5 has 1 5 10 10 5 1. Row 6 has 1 6 15 20 15 6 1. Two arrows from the 3 and 3 in row 3 point down to the 6 in row 4, illustrating Pascal's identity. Pascal's Triangle — rows n = 0 to n = 6 n = 0 1 n = 1 1 1 n = 2 1 2 1 n = 3 1 3 3 1 n = 4 1 4 6 4 1 n = 5 1 5 10 10 5 1 n = 6 1 6 15 20 15 6 1 each entry = sum of the two entries above it (Pascal's identity)
Pascal's triangle. The $n$-th row gives the coefficients of $(a+b)^n$. The arrows show Pascal's identity: $\binom{3}{1} + \binom{3}{2} = 3 + 3 = 6 = \binom{4}{2}$. Every interior entry is the sum of the two entries diagonally above it.

Each row of the triangle gives you the full expansion of (a + b)^n immediately. To expand (a + b)^6, read row 6: coefficients 1, 6, 15, 20, 15, 6, 1.

Notice the symmetry: the triangle reads the same left-to-right as right-to-left. This reflects \binom{n}{r} = \binom{n}{n-r}. The row also sums to 2^n — set a = b = 1 in the binomial theorem: (1+1)^n = \sum \binom{n}{r} = 2^n. These properties are explored in detail in binomial coefficients.

Interactive: build the expansion term by term

Drag the red point along the horizontal axis to set n from 1 to 6. The figure shows the terms of (1 + x)^n evaluated at x = 0.5, stacked as a bar chart. Each bar represents one term \binom{n}{r}(0.5)^r, and the total height is (1.5)^n.

Interactive: terms of (1+x) to the n for x equals 0.5A bar chart with r from 0 to 6 on the horizontal axis and term value on the vertical axis. For a given n, each bar shows the value of C(n,r) times 0.5 to the r. A draggable point controls n. The total height of all bars equals 1.5 to the n. r (term index) value 0 1 2 3 4 5 1 2 3 4 5 ↔ drag the red point
Each bar represents one term $\binom{n}{r}(0.5)^r$ in the expansion of $(1 + 0.5)^n = (1.5)^n$. As $n$ grows, more terms appear and their sum grows. For $n = 4$: $1 + 2 + 1.5 + 0.5 + 0.0625 = 5.0625 = (1.5)^4$.

Some useful special cases

Setting b = -b. Replace b with -b in the theorem:

(a - b)^n = \sum_{r=0}^{n} \binom{n}{r}\, a^{n-r}\, (-b)^r = \sum_{r=0}^{n} (-1)^r \binom{n}{r}\, a^{n-r}\, b^r

The terms alternate in sign: +, -, +, -, \dots

Setting a = 1, b = x. This gives the often-used form:

(1 + x)^n = \sum_{r=0}^{n} \binom{n}{r}\, x^r = 1 + nx + \binom{n}{2}x^2 + \cdots + x^n

Setting a = b = 1.

2^n = \sum_{r=0}^{n} \binom{n}{r}

The sum of all the binomial coefficients in row n is 2^n. This makes sense: 2^n is the total number of subsets of an n-element set, and \binom{n}{r} counts the subsets of size r.

Setting a = 1, b = -1.

0 = \sum_{r=0}^{n} (-1)^r \binom{n}{r}

The alternating sum of binomial coefficients is always 0. More on both of these in binomial coefficients.

Two worked examples

Example 1: Expand $(2x + 3)^5$ and find the coefficient of $x^3$

Step 1. Identify a, b, and n. Here a = 2x, b = 3, n = 5.

Why: the binomial theorem requires two terms raised to a power. Writing (2x + 3)^5 in the form (a + b)^n pins down all three ingredients.

Step 2. Write the general term.

T_{r+1} = \binom{5}{r}(2x)^{5-r}(3)^r = \binom{5}{r} \cdot 2^{5-r} \cdot 3^r \cdot x^{5-r}

Why: (2x)^{5-r} = 2^{5-r} \cdot x^{5-r} separates the numerical coefficient from the power of x.

Step 3. Find the term with x^3. Set 5 - r = 3, giving r = 2.

T_3 = \binom{5}{2} \cdot 2^3 \cdot 3^2 = 10 \cdot 8 \cdot 9 = 720

Why: r = 2 means the third term. The coefficient \binom{5}{2} = 10, and the numerical part 2^3 \cdot 3^2 = 72 multiplied by 10 gives 720.

Step 4. Write the full expansion for verification. Using r = 0, 1, 2, 3, 4, 5:

= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243

Spot-check: r = 0 gives \binom{5}{0} \cdot 2^5 \cdot 3^0 = 32. r = 5 gives \binom{5}{5} \cdot 2^0 \cdot 3^5 = 243. The first term is (2 \cdot 1)^5 \cdot 3^0 \cdot x^5 — consistent. The last term is 3^5 = 243 — also consistent with setting x = 0: (0 + 3)^5 = 243.

Result. The coefficient of x^3 in (2x + 3)^5 is 720.

The six terms of (2x+3) to the fifth power with the x cubed term highlightedSix terms are listed in a row: 32x to the 5, 240x to the 4, 720x cubed (highlighted in red), 1080x squared, 810x, and 243. Below, the components of the highlighted term are shown: C(5,2) equals 10, times 2 cubed equals 8, times 3 squared equals 9, giving 720. (2x + 3)⁵ — full expansion 32x⁵ + 240x⁴ + 720x³ + 1080x² + 810x + 243 r = 2: T₃ = C(5,2) · (2x)³ · 3² C(5,2) = 10 × 2³ = 8 × 3² = 9 coefficient of x³ = 10 × 8 × 9 = 720
The expansion of $(2x + 3)^5$ has six terms. The $x^3$ term (highlighted) corresponds to $r = 2$: the coefficient is $\binom{5}{2} \cdot 2^3 \cdot 3^2 = 10 \cdot 8 \cdot 9 = 720$.

The coefficient 720 emerges from three multiplicative pieces: the combinatorial count \binom{5}{2}, the power of 2 from the a-part, and the power of 3 from the b-part. Missing any one of these three is the most common calculation error.

Example 2: Find the $4$th term and the last term of $\left(x - \dfrac{2}{x}\right)^9$

Step 1. Identify a, b, and n. Here a = x, b = -\dfrac{2}{x}, n = 9.

Why: the negative sign belongs to b. Writing b = -2/x ensures the alternating signs appear automatically through (-1)^r.

Step 2. Write the general term.

T_{r+1} = \binom{9}{r}\, x^{9-r}\, \left(-\frac{2}{x}\right)^r = \binom{9}{r}\, (-2)^r\, x^{9-r}\, x^{-r} = \binom{9}{r}\, (-2)^r\, x^{9-2r}

Why: \left(-\frac{2}{x}\right)^r = (-2)^r \cdot x^{-r}. Combining x^{9-r} \cdot x^{-r} = x^{9-2r} simplifies the exponent.

Step 3. Find the 4th term (r = 3).

T_4 = \binom{9}{3}(-2)^3\, x^{9-6} = 84 \cdot (-8) \cdot x^3 = -672\,x^3

Why: \binom{9}{3} = \frac{9 \cdot 8 \cdot 7}{3!} = 84, and (-2)^3 = -8. The negative sign is automatic.

Step 4. Find the last term (r = 9).

T_{10} = \binom{9}{9}(-2)^9\, x^{9-18} = 1 \cdot (-512) \cdot x^{-9} = -\frac{512}{x^9}

Spot-check: the expansion has 9 + 1 = 10 terms, so T_{10} is indeed the last. Its sign is negative because (-2)^9 = -512 (odd power).

Result. The 4th term is -672\,x^3 and the last term is -\dfrac{512}{x^9}.

The 4th and last terms of the expansion of (x minus 2 over x) to the 9A horizontal timeline of terms T1 through T10. T4 and T10 are highlighted. T4 is shown to equal negative 672 x cubed and T10 equals negative 512 over x to the 9. The general term formula T(r+1) = C(9,r) times negative 2 to the r times x to the 9 minus 2r is shown above. (x − 2/x)⁹ — 10 terms total T(r+1) = C(9,r) · (−2)ʳ · x^(9−2r) T₁ T₂ T₃ T₄ T₅ T₆ T₇ T₈ T₉ T₁₀ −672x³ C(9,3)·(−8)·x³ −512/x⁹ C(9,9)·(−512)·x⁻⁹ signs alternate: +, −, +, −, ... because of (−2)ʳ
The $4$th term ($r = 3$) and the last term ($r = 9$) of $(x - 2/x)^9$. The sign of each term depends on whether $r$ is even or odd, since $(-2)^r$ alternates.

The power of x in the general term is 9 - 2r, which decreases as r increases: x^9, x^7, x^5, x^3, x^1, x^{-1}, x^{-3}, x^{-5}, x^{-7}, x^{-9}. The general and middle terms article takes this further — finding the term independent of x, the middle term, and the term with the greatest coefficient.

Common confusions

Going deeper

If you can expand (a + b)^n, find any specific term, and read coefficients off Pascal's triangle, you have the full toolkit for problems on the binomial theorem. The rest of this section is for readers who want to see a few more connections.

The binomial theorem for (1+x)^n and approximations

When x is small compared to 1, the higher-power terms of (1 + x)^n become negligible. The first two terms give the linear approximation:

(1 + x)^n \approx 1 + nx \qquad \text{for small } x

For instance, (1.01)^{10} \approx 1 + 10(0.01) = 1.1. The exact value is 1.10462..., so the approximation is accurate to about 4\%. Including the x^2 term improves it: 1 + 0.1 + \binom{10}{2}(0.01)^2 = 1 + 0.1 + 0.0045 = 1.1045, matching four decimal places.

This approximation is used throughout physics and engineering for quick estimates — and it is the seed of the general binomial series for non-integer exponents.

Connection to the multinomial theorem

The binomial theorem expands (a + b)^n — two terms raised to a power. The natural generalisation is the multinomial theorem, which expands (a_1 + a_2 + \cdots + a_k)^n. The coefficients become multinomial coefficients \frac{n!}{r_1!\, r_2! \cdots r_k!}, and the sum runs over all non-negative integer sequences (r_1, r_2, \dots, r_k) with r_1 + r_2 + \cdots + r_k = n.

Pingala and the Indian roots of Pascal's triangle

The triangular array of binomial coefficients is often called Pascal's triangle in textbooks, but the pattern was known in India long before. Pingala's Chandahshastra (around 200 BCE) described the systematic enumeration of binary patterns in Sanskrit prosody — effectively constructing the rows of the triangle. Halayudha's 10th-century commentary made the additive rule (\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}) explicit. The triangle is sometimes called the Meru-prastara (staircase of Mount Meru) in Indian mathematical tradition.

Where this leads next