In short

For any rational number n (fractional, negative, or otherwise non-positive-integer) and |x| < 1, the generalised binomial theorem gives (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots, an infinite series. Unlike the positive-integer case, the expansion never terminates, and it is valid only when |x| < 1. This series is the basis for approximating expressions like \sqrt{1.02} or \frac{1}{(1+x)^3} without a calculator.

A civil engineer needs to estimate \sqrt{104} during a quick site calculation. She rewrites it as \sqrt{100 \cdot 1.04} = 10\sqrt{1.04} = 10(1 + 0.04)^{1/2}. If the binomial theorem worked for n = 1/2, she could expand (1 + 0.04)^{1/2} and keep a few terms for a fast approximation. Does it work?

The answer is yes — with a condition. When the exponent n is a positive integer, (a + b)^n expands into finitely many terms and the expansion is exact for all values of a and b. When n is a fraction or a negative number, the expansion becomes an infinite series and it converges only when |x| < 1 (in the standard form (1+x)^n). Respecting this convergence condition is the price of admission to a much larger world of expansions.

Why does the finite expansion break down?

Recall from the positive-integer theorem that the general term involves \binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}. When n is a positive integer and r > n, one of the factors in the numerator is (n - n) = 0, which kills every term past T_{n+1}. That is why the expansion terminates at n + 1 terms.

When n = 1/2, the factors in the numerator are \frac{1}{2}, \frac{1}{2} - 1, \frac{1}{2} - 2, \frac{1}{2} - 3, \dots = \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, -\frac{5}{2}, \dots. None of these is ever zero. The product never vanishes, so no term is zero, and the series goes on forever.

Why the expansion terminates for positive integer n but not for rational nTwo rows show the numerator factors for n=4 and n=1/2. For n=4, the factors are 4, 3, 2, 1, 0 — the zero kills all subsequent terms. For n=1/2, the factors are 1/2, -1/2, -3/2, -5/2, -7/2 — none is zero, so the series continues forever. Why does the series terminate — or not? n = 4: 4 · 3 · 2 · 1 · 0 · … ← kills all later terms n = ½: ½ · −½ · −³⁄₂ · −⁵⁄₂ · −⁷⁄₂ · … ← never zero → infinite series Numerator factors of the general term: n, (n−1), (n−2), (n−3), … A positive integer n makes one factor zero; a rational n never does
For $n = 4$, the numerator factor hits zero at the fifth step, terminating the expansion. For $n = 1/2$, the factors $\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \dots$ are never zero, producing an infinite series.

The theorem

Binomial Theorem for Rational Index

For any n \in \mathbb{Q} (where n is not a non-negative integer) and |x| < 1:

(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots = \sum_{r=0}^{\infty} \frac{n(n-1)\cdots(n-r+1)}{r!}\,x^r

The coefficient of x^r is written \binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}, the generalised binomial coefficient. The series converges if and only if |x| < 1.

Three things change from the positive-integer version:

  1. The series is infinite. There is no last term.
  2. Convergence requires |x| < 1. Outside this range, the partial sums diverge.
  3. The base must be (1 + x). You cannot directly expand (a + b)^n for arbitrary a, b. Instead, factor out: (a + b)^n = a^n\left(1 + \frac{b}{a}\right)^n, and apply the theorem to x = b/a provided |b/a| < 1, i.e., |b| < |a|.
Comparison of the binomial theorem for positive integer vs rational exponentTwo columns. Left: positive integer n, finite terms, valid for all x. Right: rational n, infinite series, valid only for absolute value of x less than 1. Positive integer n vs. rational n n ∈ ℤ⁺ n ∈ ℚ \ ℤ⁺ Finite (n + 1 terms) Infinite series Valid for all x Valid only for |x| < 1 Coefficients: C(n, r) Generalised C(n, r) Exact identity Convergent series when n is a positive integer, both versions agree (and the series terminates)
The key differences between the binomial theorem for positive integer exponent and for rational exponent. The rational version trades finiteness for generality.

Important expansions

Several specific values of n arise repeatedly. Working out the first few terms by hand builds fluency.

n = -1: the geometric series.

(1 + x)^{-1} = 1 - x + x^2 - x^3 + x^4 - \cdots \qquad (|x| < 1)

Check: the coefficient of x^r is \binom{-1}{r} = \frac{(-1)(-2)(-3)\cdots(-r)}{r!} = \frac{(-1)^r \cdot r!}{r!} = (-1)^r.

This is exactly the sum of the infinite geometric series \frac{1}{1+x} = \sum_{r=0}^{\infty} (-x)^r. The binomial theorem and the GP formula agree perfectly.

n = -2:

(1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 - \cdots \qquad (|x| < 1)

The coefficient of x^r is (-1)^r(r+1). This is the term-by-term derivative of (1+x)^{-1}, divided by -1 — a connection that becomes clear once you study calculus.

n = 1/2: the square root.

(1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots \qquad (|x| < 1)

The coefficients are \frac{1}{2}, \frac{\frac{1}{2} \cdot (-\frac{1}{2})}{2!} = -\frac{1}{8}, \frac{\frac{1}{2} \cdot (-\frac{1}{2}) \cdot (-\frac{3}{2})}{3!} = \frac{1}{16}, and so on.

n = -1/2: the inverse square root.

(1 + x)^{-1/2} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \cdots \qquad (|x| < 1)
Four important binomial expansions for rational nFour rows showing the first few terms of (1+x) to the power -1, -2, 1/2, and -1/2. Each row lists the value of n and the expansion up to the x-cubed term. Four expansions to remember n = −1 1 − x + x² − x³ + ⋯ n = −2 1 − 2x + 3x² − 4x³ + ⋯ n = ½ 1 + ½x − ⅛x² + ¹⁄₁₆x³ − ⋯ n = −½ 1 − ½x + ³⁄₈x² − ⁵⁄₁₆x³ + ⋯ all valid for |x| < 1 n = −1 recovers the geometric series; n = ½ gives square roots
The four most commonly used binomial expansions for rational $n$. The $n = -1$ case is the infinite geometric series. Each is valid only for $|x| < 1$.

Applications in approximation

The real power of this theorem is computation. When x is small, the first two or three terms of (1 + x)^n give an excellent approximation.

Linear approximation: (1 + x)^n \approx 1 + nx for small |x|.

Quadratic approximation: (1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 for small |x|.

Take \sqrt{1.02} = (1 + 0.02)^{1/2}. Here x = 0.02 and n = 1/2.

The quadratic approximation matches to five decimal places — from a two-term calculation.

Graph comparing sqrt(1+x) with its linear and quadratic binomial approximationsA graph on the domain x from -0.5 to 0.5. Three curves are plotted: the exact function sqrt(1+x), the linear approximation 1 + x/2, and the quadratic approximation 1 + x/2 - x-squared/8. Near x=0 all three nearly overlap. As x grows, the linear approximation diverges first. x y 0 0.5 −0.5 √(1+x) 1 + x/2 quadratic x = 0 near x = 0, all three curves nearly coincide
The exact function $\sqrt{1+x}$ (solid), the linear approximation $1 + x/2$ (dashed), and the quadratic approximation $1 + x/2 - x^2/8$ (dotted red). Near $x = 0$, the quadratic approximation hugs the true curve closely. As $|x|$ grows toward $1$, the gap widens.

Interactive: see the approximation improve term by term

Drag the red point to change the number of terms kept in the expansion of (1 + x)^{1/2} (from 1 term up to 6 terms). The curve shows how the partial sum compares to the exact \sqrt{1+x} on the interval (-0.8, 0.8).

Interactive: partial sums of the binomial series for square root of 1+xA graph showing the exact function sqrt(1+x) and a partial sum approximation. A draggable point controls the number of terms from 1 to 6. As more terms are added, the partial sum curve approaches the exact curve more closely on the interval negative 0.8 to 0.8. x y 0 −1 1 ↔ drag the red point
The solid curve is $\sqrt{1+x}$. The dashed red curve is the partial sum of the binomial series. As you increase the number of terms, the approximation improves — rapidly near $x = 0$, more slowly near $|x| = 1$.

Two worked examples

Example 1: Approximate $\sqrt[3]{1001}$ correct to five decimal places

Step 1. Rewrite in the form (1 + x)^n with small x.

\sqrt[3]{1001} = \sqrt[3]{1000 \cdot 1.001} = 10 \cdot (1 + 0.001)^{1/3}

Here n = 1/3 and x = 0.001.

Why: factor out the nearest perfect cube (1000 = 10^3) to make x as small as possible. Smaller x means fewer terms needed for accuracy.

Step 2. Write the first three terms of the expansion.

(1 + x)^{1/3} = 1 + \frac{1}{3}x + \frac{\frac{1}{3}\left(\frac{1}{3} - 1\right)}{2!}x^2 + \cdots = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \cdots

Why: the second coefficient is \frac{(1/3)(1/3 - 1)}{2} = \frac{(1/3)(-2/3)}{2} = -\frac{1}{9}.

Step 3. Substitute x = 0.001.

= 1 + \frac{0.001}{3} - \frac{(0.001)^2}{9} + \cdots = 1 + 0.000\,333\,33\ldots - 0.000\,000\,111\ldots + \cdots

Why: the x^2 term is already of order 10^{-7}, contributing only at the seventh decimal place. For five-decimal-place accuracy, two terms suffice.

Step 4. Compute the result.

\sqrt[3]{1001} = 10 \times 1.000\,333\,22\ldots \approx 10.003\,332\,2

A calculator gives \sqrt[3]{1001} = 10.003\,332\,22\ldots, confirming five-decimal-place accuracy.

Result: \sqrt[3]{1001} \approx 10.00333.

Breaking down the approximation of the cube root of 1001A diagram showing the decomposition: cube root of 1001 = 10 times (1 + 0.001) to the 1/3. The first three terms of the expansion are listed: 1, plus 0.000333, minus 0.000000111. The sum 10.00333 is highlighted. ∛1001 = 10 × (1 + 0.001)^(1/3) Term 1: 1 = C(⅓, 0) · (0.001)⁰ Term 2: + 0.000 333 3… = ⅓ × 0.001 Term 3: − 0.000 000 111… = −⅑ × (0.001)² × 10 → 10.003 33 exact: 10.003 332 22… — five-place accuracy from two terms
The expansion of $(1 + 0.001)^{1/3}$. The first correction term $0.001/3 \approx 0.000333$ does the heavy lifting. The $x^2$ term is negligible at order $10^{-7}$. Multiplying by $10$ gives $\sqrt[3]{1001}$ to five decimal places.

The strategy is always the same: extract the largest perfect power, make x small, and keep enough terms for the desired precision. The smaller x is, the faster the series converges.

Example 2: Expand $\dfrac{1}{(1 - 2x)^3}$ up to the $x^3$ term

Step 1. Rewrite in the form (1 + u)^n.

\frac{1}{(1 - 2x)^3} = (1 + (-2x))^{-3}

Here n = -3 and u = -2x, valid for |-2x| < 1, i.e., |x| < \frac{1}{2}.

Why: replacing x with -2x in (1 + x)^n means the convergence condition becomes |{-2x}| < 1.

Step 2. Write the general coefficient. For (1 + u)^{-3}, the coefficient of u^r is:

\binom{-3}{r} = \frac{(-3)(-4)(-5)\cdots(-3-r+1)}{r!} = (-1)^r \frac{(r+1)(r+2)}{2}

Why: the numerator has r factors from -3 down: (-3)(-4)\cdots(-(r+2)). The absolute value is \frac{(r+2)!}{2!}, and the sign is (-1)^r.

Step 3. Substitute u = -2x and expand.

(1 - 2x)^{-3} = \sum_{r=0}^{\infty} (-1)^r \frac{(r+1)(r+2)}{2} (-2x)^r = \sum_{r=0}^{\infty} \frac{(r+1)(r+2)}{2} \cdot 2^r \cdot x^r

The first four terms (r = 0, 1, 2, 3):

  • r = 0: \frac{1 \cdot 2}{2} \cdot 1 = 1
  • r = 1: \frac{2 \cdot 3}{2} \cdot 2 = 6
  • r = 2: \frac{3 \cdot 4}{2} \cdot 4 = 24
  • r = 3: \frac{4 \cdot 5}{2} \cdot 8 = 80

Why: (-1)^r \cdot (-2)^r = (-1)^r \cdot (-1)^r \cdot 2^r = 2^r (the two signs cancel).

Step 4. Write the result.

\frac{1}{(1 - 2x)^3} = 1 + 6x + 24x^2 + 80x^3 + \cdots \qquad \left(|x| < \frac{1}{2}\right)

Spot-check at x = 0.1: \frac{1}{(0.8)^3} = \frac{1}{0.512} = 1.953125. Partial sum: 1 + 0.6 + 0.24 + 0.08 = 1.92. Close, and the remaining terms will bring it to 1.953\dots

Result: \dfrac{1}{(1 - 2x)^3} = 1 + 6x + 24x^2 + 80x^3 + \cdots for |x| < \dfrac{1}{2}.

Coefficients of the expansion of 1 over (1 minus 2x) cubedFour terms of the expansion are shown: 1, 6x, 24x squared, 80x cubed. Below each, the computation is shown: C(-3,r) times (-2x) to the r simplified. The convergence condition |x| less than 1/2 is noted. 1/(1 − 2x)³ = (1 + (−2x))⁻³ 1 + 6x + 24x² + 80x³ + ⋯ r = 0 r = 1 r = 2 r = 3 1·2/2 × 2⁰ 2·3/2 × 2¹ 3·4/2 × 2² 4·5/2 × 2³ coefficient of xʳ = (r+1)(r+2)/2 × 2ʳ convergence: |x| < ½ check: at x = 0.1, partial sum 1.92 ≈ 1/(0.8)³ = 1.953…
The expansion of $(1 - 2x)^{-3}$. The coefficient pattern $(r+1)(r+2)/2 \cdot 2^r$ grows rapidly, but convergence is guaranteed as long as $|x| < 1/2$. The spot-check at $x = 0.1$ confirms the series is heading toward the exact value.

Notice the clean pattern in the coefficients: 1, 6, 24, 80. The general coefficient (r+1)(r+2) \cdot 2^{r-1} blends a combinatorial factor with a geometric growth — a signature of negative-exponent binomial expansions.

Common confusions

Going deeper

If you can expand (1 + x)^n for rational n, compute generalised binomial coefficients, and use the series for approximation, you have the full toolkit for this chapter. What follows is for readers who want the rigorous backdrop.

Convergence at the boundary |x| = 1

The condition |x| < 1 guarantees convergence, but what happens at x = 1 and x = -1?

The full boundary behaviour is governed by Abel's theorem and is studied in analysis courses.

Connection to Taylor series

The binomial series for (1 + x)^n is, in fact, the Taylor series of (1 + x)^n about x = 0. The r-th derivative of (1+x)^n at x = 0 is n(n-1)(n-2)\cdots(n-r+1), so the Taylor coefficient is \frac{n(n-1)\cdots(n-r+1)}{r!} — exactly the generalised binomial coefficient. The radius of convergence |x| < 1 comes from the ratio test applied to consecutive terms.

Historical note

The extension of the binomial theorem to non-integer exponents was one of the early triumphs of infinite series. The insight that (1 + x)^{1/2} could be expanded as an infinite sum, and that this sum actually converged to \sqrt{1+x}, opened the door to the systematic use of power series in mathematics — an idea that eventually led to the full theory of Taylor and Maclaurin series.

Where this leads next