P1.4.1 - Binomial expansion of (a + bx)^n

P1.4.1 - Binomial expansion of (a + bx)^n

The binomial expansion is a controlled way to expand a bracket with two terms. For positive integer powers it gives an exact finite polynomial. For rational powers, such as square roots or reciprocal powers, it gives an infinite series that can be used for approximation, but only inside a stated range of validity.

Positive integer powers and coefficients

A binomial is an expression with two terms, such as a + bx. When n is a positive integer, (a+bx)^n means multiplying n identical brackets:

(a+bx)n=(a+bx)(a+bx)⋯(a+bx).(a+bx)^n=(a+bx)(a+bx)\cdots(a+bx).

Each term in the expansion comes from choosing some brackets to contribute bx and the remaining brackets to contribute a. If exactly r of the n brackets contribute bx, then the term contains (bx)^r and a^{n-r}.

Binomial coefficient

For integers n >= 0 and 0 <= r <= n,

(nr)=nCr=nCr=n!r!(n−r)!.\binom{n}{r}={}^{n}C_r=nC_r=\frac{n!}{r!(n-r)!}.

It counts the number of ways to choose r objects from n objects.

The factorial notation means

n!=n(n−1)(n−2)⋯2⋅1,0!=1.n! = n(n-1)(n-2)\cdots 2\cdot 1,\qquad 0! = 1.

That same nCr notation appears again in binomial probabilities in statistics, because binomial probability also counts how many ways r successes can occur in n trials. In this lesson, we use the coefficient notation for algebraic expansion.

Binomial expansion for positive integer n

For positive integer n,

(a+bx)n=an+(n1)an−1(bx)+(n2)an−2(bx)2+⋯+(nr)an−r(bx)r+⋯+(bx)n.(a+bx)^n=a^n+\binom{n}{1}a^{n-1}(bx)+\binom{n}{2}a^{n-2}(bx)^2+\cdots+\binom{n}{r}a^{n-r}(bx)^r+\cdots+(bx)^n.

The coefficient pattern can also be read from Pascal's triangle. For example:

nCoefficients in (p+q)^n
01
11, 1
21, 2, 1
31, 3, 3, 1
41, 4, 6, 4, 1
51, 5, 10, 10, 5, 1

There are two useful relations between the coefficients:

(nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r}

and

(nr)=(n−1r−1)+(n−1r).\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}.

The first says the row is symmetric. The second is the rule that creates the next row of Pascal's triangle by adding adjacent entries.

Worked example: expand (2-3x)^4.

Use the n=4 row of Pascal's triangle:

1, 4, 6, 4, 1.1,\ 4,\ 6,\ 4,\ 1.

Then

(2−3x)4=1(2)4+4(2)3(−3x)+6(2)2(−3x)2+4(2)(−3x)3+(−3x)4=16−96x+216x2−216x3+81x4.\begin{aligned} (2-3x)^4 &=1(2)^4+4(2)^3(-3x)+6(2)^2(-3x)^2+4(2)(-3x)^3+(-3x)^4\\ &=16-96x+216x^2-216x^3+81x^4. \end{aligned}

Notice the brackets around (-3x). They are not decoration: without them, the signs and powers are easy to lose.

Finding a particular term efficiently

You do not always need the whole expansion. The general term tells you exactly where each power of x comes from.

For (a+bx)^n, the term containing x^r is

(nr)an−r(bx)r.\binom{n}{r}a^{n-r}(bx)^r.

So the coefficient of x^r is

(nr)an−rbr.\binom{n}{r}a^{n-r}b^r.

This is often faster and safer than expanding a long expression. It also helps when the question asks for a particular coefficient or a missing constant.

Worked example: find the term in x^3 in the expansion of (5-2x)^6.

The power x^3 occurs when r=3.

(63)56−3(−2x)3=20⋅125⋅(−8)x3.\binom{6}{3}5^{6-3}(-2x)^3 =20\cdot 125\cdot (-8)x^3.

Therefore the term in x^3 is

−20000x3.-20000x^3.

If the question asks for the coefficient, the answer is -20000. If it asks for the term, include the x^3.

Finding a particular term efficiently Continued

A common trap is to make r match the term number instead of the power of x. The first term has r=0, the second term has r=1, and so on. The term containing x^4 has r=4, not r=5.

Rational powers and infinite series

When the power is not a positive integer, the expansion usually does not stop. Edexcel p1.4.1 extends the binomial expansion to rational powers such as 1/2, -1, and -3/2.

The standard form is (1+u)^n, where u is the expression being raised to increasing powers.

General binomial expansion

For rational n,

(1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯ ,(1+u)^n =1+nu+\frac{n(n-1)}{2!}u^2+\frac{n(n-1)(n-2)}{3!}u^3+\cdots,

valid when |u| < 1.

For (a+bx)^n, first factor out a:

(a+bx)n=an(1+bax)n.(a+bx)^n=a^n\left(1+\frac{b}{a}x\right)^n.

Then use

u=bax.u=\frac{b}{a}x.

This is the main decision in rational binomial expansion. The expression inside the general formula is not automatically x; it is whatever replaces u.

Worked example: find the expansion of (4-3x)^{1/2} up to and including the term in x^3.

First put the expression into the form a^n(1+u)^n:

(4−3x)1/2=41/2(1−3x4)1/2=2(1−3x4)1/2.(4-3x)^{1/2}=4^{1/2}\left(1-\frac{3x}{4}\right)^{1/2} =2\left(1-\frac{3x}{4}\right)^{1/2}.

Here

n=12,u=−3x4.n=\frac12,\qquad u=-\frac{3x}{4}.

Use the first four terms of the general expansion:

(1+u)1/2=1+12u+12(12−1)2!u2+12(12−1)(12−2)3!u3+⋯ .(1+u)^{1/2} =1+\frac12u+\frac{\frac12\left(\frac12-1\right)}{2!}u^2 +\frac{\frac12\left(\frac12-1\right)\left(\frac12-2\right)}{3!}u^3+\cdots.

Simplify the coefficients:

(1+u)1/2=1+12u−18u2+116u3+⋯ .(1+u)^{1/2}=1+\frac12u-\frac18u^2+\frac{1}{16}u^3+\cdots.

Now substitute u=-3x/4 and multiply by the outside factor 2:

(4−3x)1/2=2[1+12(−3x4)−18(−3x4)2+116(−3x4)3+⋯ ]=2−34x−964x2−27512x3+⋯ .\begin{aligned} (4-3x)^{1/2} &=2\left[1+\frac12\left(-\frac{3x}{4}\right)-\frac18\left(-\frac{3x}{4}\right)^2+\frac{1}{16}\left(-\frac{3x}{4}\right)^3+\cdots\right]\\ &=2-\frac34x-\frac{9}{64}x^2-\frac{27}{512}x^3+\cdots. \end{aligned}

There are two places where errors commonly happen: forgetting the outside factor 2, and not squaring or cubing the whole bracketed expression (-3x/4).

Validity and approximation

For positive integer n, the binomial expansion is a finite polynomial, so there is no convergence condition to state. For rational non-integer n, the expansion is an infinite series, and Edexcel expects you to know its range of validity.

Since

(a+bx)n=an(1+bax)n,(a+bx)^n=a^n\left(1+\frac{b}{a}x\right)^n,

the rational-index expansion is valid when

∣bax∣<1.\left|\frac{b}{a}x\right|<1.

Equivalently,

∣x∣<∣ab∣|x|<\left|\frac{a}{b}\right|

when b != 0.

This condition is not just a technical footnote. If a question asks whether a proposed substitution is valid, you must state the range and compare the value with the range. A vague sentence such as "it is outside the range" is usually not enough unless the range has already been made clear.

Worked example: use the binomial expansion to approximate sqrt(3).

Start from

4−x=(4−x)1/2=2(1−x4)1/2.\sqrt{4-x}=(4-x)^{1/2}=2\left(1-\frac{x}{4}\right)^{1/2}.

Using the general expansion,

(4−x)1/2=2[1−x8−x2128−x31024+⋯ ]=2−x4−x264−x3512+⋯ .\begin{aligned} (4-x)^{1/2} &=2\left[1-\frac{x}{8}-\frac{x^2}{128}-\frac{x^3}{1024}+\cdots\right]\\ &=2-\frac{x}{4}-\frac{x^2}{64}-\frac{x^3}{512}+\cdots. \end{aligned}

To approximate sqrt(3), choose x=1, because 4-x=3.

The validity check is

∣−x4∣<1.\left|-\frac{x}{4}\right|<1.

At x=1, this gives 1/4 < 1, so the substitution is valid.

Now substitute:

3≈2−14−164−1512=887512=1.732421875.\sqrt{3}\approx 2-\frac14-\frac{1}{64}-\frac{1}{512} =\frac{887}{512} =1.732421875.

The approximation works because the powers of x/4 get smaller. In general, the closer u=(b/a)x is to 0, the faster the early terms settle down.

Expanding rational functions after decomposition

The specification notes that binomial expansion may be used after decomposing a rational function into partial fractions. The binomial step begins once each part has been written as a constant times (1+cx)^n.

For this lesson, the key skill is not the decomposition itself. The key skill is to expand each simple part correctly and then combine the validity conditions.

Worked example: expand

f(x)=52+x−31−2xf(x)=\frac{5}{2+x}-\frac{3}{1-2x}

up to and including the term in x^2, and state the range of validity.

Start with the first fraction:

52+x=52(1+x2)−1.\frac{5}{2+x}=\frac52\left(1+\frac{x}{2}\right)^{-1}.

Using (1+u)^{-1}=1-u+u^2-\cdots,

52+x=52[1−x2+(x2)2+⋯ ]=52−54x+58x2+⋯ .\frac{5}{2+x} =\frac52\left[1-\frac{x}{2}+\left(\frac{x}{2}\right)^2+\cdots\right] =\frac52-\frac54x+\frac58x^2+\cdots.

This part is valid when

∣x2∣<1,\left|\frac{x}{2}\right|<1,

so |x|<2.

Now expand the second fraction:

−31−2x=−3(1−2x)−1.-\frac{3}{1-2x}=-3(1-2x)^{-1}.

Using u=-2x,

−3(1−2x)−1=−3[1+2x+4x2+⋯ ]=−3−6x−12x2+⋯ .-3(1-2x)^{-1} =-3[1+2x+4x^2+\cdots] =-3-6x-12x^2+\cdots.

This part is valid when |-2x|<1, so |x|<1/2.

Combine the two expansions:

f(x)=(52−54x+58x2)+(−3−6x−12x2)+⋯=−12−294x−918x2+⋯ .\begin{aligned} f(x) &=\left(\frac52-\frac54x+\frac58x^2\right)+\left(-3-6x-12x^2\right)+\cdots\\ &=-\frac12-\frac{29}{4}x-\frac{91}{8}x^2+\cdots. \end{aligned}

Both series must be valid at the same time, so the overall range is the intersection:

∣x∣<12.|x|<\frac12.

Exam strategy and common traps

The most reliable binomial-expansion routine is:

  1. Decide whether the power is a positive integer or a rational non-integer.
  2. If the power is a positive integer, use the finite expansion or the general term.
  3. If the power is rational, rewrite the bracket as a^n(1+u)^n.
  4. Expand using powers of the whole expression u.
  5. Multiply every term by the outside factor a^n.
  6. If the expansion is infinite, state the validity condition |u|<1 when asked, and compare any substituted value with it.

Here is the key contrast.

SituationExpansion typeValidity comment
n is a positive integerfinite polynomialno convergence condition
n is rational but not a positive integerinfinite seriesvalid when `

Several errors are predictable:

  • using |x|<1 instead of finding the actual u;
  • dropping the outside factor after factorising;
  • writing powers of x instead of powers of the whole bracketed expression;
  • using decimal coefficients when exact fractions are clearer;
  • giving a validity conclusion without stating the range.

For rational powers, the binomial expansion is not really about x; it is about the inner expression u. Find u, expand in powers of u, and use |u|<1.