Drop the requirement that n is a positive integer, and the finite triangle of coefficients unfolds into an infinite series — valid only within a strict window.
9 min read3 worked examples
Extending the binomial expansion beyond positive integers
For a positive integer $n$, the binomial expansion of $(1+x)^n$ is a finite sum of $n+1$ terms, with coefficients $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ — this relies on the factorial $n!$ making sense, which requires $n$ to be a non-negative integer. When $n$ is instead any rational number (a fraction, or negative, or both), $n!$ is no longer defined, so the coefficient formula must be rewritten without factorials:
Each coefficient's numerator is a product of $r$ consecutive descending factors starting from $n$ itself — this works perfectly well even when $n$ is a fraction or negative, since it never actually requires computing a factorial of $n$.
The critical difference: an infinite series
When $n$ is a positive integer, the numerator $n(n-1)(n-2)\cdots$ eventually hits a factor of zero (at $r=n+1$), so the expansion terminates naturally after $n+1$ terms. For any other rational $n$, this never happens — the expansion is an infinite series that never terminates, and it is only valid (converges to the true value of $(1+x)^n$) when
$|x| < 1$
Outside this range, the series diverges and no longer represents $(1+x)^n$ at all. This condition must always be stated alongside the expansion — it is not optional, and questions frequently award marks specifically for stating it correctly.
Approximating values with a few terms
Because the series converges (for $|x|<1$), taking just the first few terms gives a good numerical approximation to $(1+x)^n$, provided $x$ is small — the smaller $|x|$ is, the faster the later terms shrink towards zero and the better the approximation becomes using only a handful of terms.
Worked example 1
Expanding $(1+x)^n$ for a fractional n
Find the binomial expansion of $(1+x)^{1/2}$ up to and including the term in $x^3$, and state the range of values of $x$ for which the expansion is valid.
MethodSubstitute $n=\frac12$ into the general expansion, term by term
1
$x^1$ term: $nx = \dfrac12 x$.
Direct substitution of $n=\frac12$ into the linear term.
$n=\frac12$ is not a positive integer, so the series is infinite and requires this convergence condition.
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Examiner's eye
Stating "valid for $|x|<1$" is a required part of a complete answer whenever $n$ is not a positive integer — omitting it, even with a perfectly correct expansion, typically loses a mark.
Worked example 2
Using the expansion to approximate a square root
Using the first three terms of the expansion of $(1+x)^{1/2}$ from Example 1, find an approximate value for $\sqrt{1.02}$.
MethodMatch 1.02 to the form 1+x, then substitute into the expansion
1
$\sqrt{1.02} = (1.02)^{1/2} = (1+0.02)^{1/2}$, so $x=0.02$.
Match the target value to the form $(1+x)^{1/2}$ used in the expansion.
2
Since $|0.02|<1$, the expansion is valid here.
Confirms the approximation is legitimate before using it.
3
Using the first three terms: $(1+0.02)^{1/2} \approx 1+\dfrac12(0.02)-\dfrac18(0.02)^2$.
Substitute $x=0.02$ into $1+\frac12x-\frac18x^2$ from Example 1.
This closely matches the true value $\sqrt{1.02}=1.009950\ldots$
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Strategy
The technique works because $x=0.02$ is small: each successive term is roughly $\frac{1}{50}$ the size of the one before, so just a few terms already give excellent accuracy — this is precisely why the method is useful for hand-calculating roots close to 1.
Worked example 3
Finding a single coefficient without expanding every term
Find the coefficient of $x^4$ in the binomial expansion of $(1-x)^{1/3}$.
MethodWrite $(1-x)^{1/3}$ as $(1+u)^n$ with $u=-x$, then apply the general term directly at $r=4$
1
$(1-x)^{1/3} = (1+u)^{1/3}$ with $u=-x$, so the coefficient of $x^4$ equals the coefficient of $u^4$ (since $u^4=(-x)^4=x^4$).
Recognise $-x$ as playing the role of $u$ in the standard expansion.
2
The general term is $\dbinom{n}{4}u^4 = \dfrac{n(n-1)(n-2)(n-3)}{4!}u^4$, with $n=\dfrac13$.
Apply the general-term formula directly at $r=4$, without writing out $r=1,2,3$ first.
Multiply the four descending factors, then divide by $4!=24$.
4
The coefficient of $x^4$ is $-\dfrac{10}{243}$.
Since $u^4=x^4$ exactly, no sign change is needed when converting back to $x$.
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A shortcut worth knowing
When a question asks for a single, specific coefficient rather than "the expansion up to $x^k$," the general term formula can be applied directly at that one value of $r$ — there's no need to generate every earlier term first. This saves significant time whenever the target power is $x^4$ or higher.
Common mistake
Treating $(1-x)^n$ as if it needed no adjustment from $(1+x)^n$ — every occurrence of $x$ in the standard expansion must be replaced by $-x$, which flips the sign of every odd-power term. Writing out $u=-x$ explicitly before substituting, as in Example 3, avoids sign slips.
Check yourself
Find the binomial expansion of $(1-x)^{-1/2}$ up to and including the term in $x^2$, and state the range of values of $x$ for which the expansion is valid. (Hint: this is $(1+u)^n$ with $u=-x$ and $n=-\frac12$.)
Method
1
With $u=-x$, $n=-\dfrac12$: $x^1$ term is $nu = \left(-\dfrac12\right)(-x) = \dfrac12x$.