Mathematics 9709 · Pure Mathematics 3

Binomial expansion for rational n

The binomial expansion no longer needs a whole-number power. It still works for any rational n, with two changes.
The series never stops. And it only works when $|x|<1$ .
In return you can now expand reciprocals (1 over something) and surds.

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots$ for rational $n$ , valid when $|x|<1$ . Use the falling-factorial top $n(n-1)(n-2)\cdots$ , never $\binom{n}{r}$ (that only works for whole numbers).

The general formula

$(1+x)^n=1+nx+\tfrac{n(n-1)}{2!}x^2+\tfrac{n(n-1)(n-2)}{3!}x^3+\cdots\quad(|x|<1)$

For each new term, multiply the last factor on top by the next number down from $n$ , then divide by the next factorial.
A square root is the power $n=\tfrac12$ , a cube root is $n=\tfrac13$ , and 1 over something is $n=-1$ .
If the inside is $kx$ , the range gets smaller: $|kx|<1$ , which means $|x|<\tfrac1{|k|}$ .

Why $|x|<1$ ? The expansion is now an infinite sum. Like a geometric series, it only settles on one finite value while the terms keep getting smaller. For that the inside has to be under 1 in size. Go past that and the running total grows without limit.

See the validity range

Drag $x$ across the line $x=1$ . Inside the band the blue running totals settle onto the gold true value of $(1+x)^n$ . Push $|x|$ to 1 or past it and they scatter.
That is the validity condition the mark scheme wants you to write down.

x = 0.50·true value 0.667·gap after 9 terms 0.001

The link to geometric series

Take $n=-1$ . The expansion is $(1+x)^{-1}=1-x+x^2-x^3+\cdots$ , a geometric series with ratio $-x$ . A geometric series adds up to a finite value exactly when $|\text{ratio}|<1$ , which here is $|x|<1$ . The general binomial does the same thing for every rational $n$ .

Worked examples

Worked example

Expand $(1+x)^{-2}$ up to $x^3$ , with validity

With $n=-2$ the falling-factorial top runs $-2,\,-3,\,-4$ : $1+(-2)x+\tfrac{(-2)(-3)}{2}x^2+\tfrac{(-2)(-3)(-4)}{6}x^3$ .
Simplify each coefficient: $-2,\ \tfrac{6}{2}=3,\ \tfrac{-24}{6}=-4$ .

Answer

1-2x+3x^2-4x^3+\cdots,\quad |x|<1

Worked example

Textbook: expand $(1-3x)^{-2}$ up to $x^3$ , with validity

Use $n=-2$ and replace $x$ by $-3x$ : $1+(-2)(-3x)+\tfrac{(-2)(-3)}{2!}(-3x)^2+\tfrac{(-2)(-3)(-4)}{3!}(-3x)^3$ .
Each $(-3x)^k$ brings a power of $3$ : $6x,\ 3(9x^2),\ -4(-27x^3)$ , i.e. $6x,\ 27x^2,\ 108x^3$ .
Validity comes from the inside: $|-3x|<1$ , so $|x|<\tfrac13$ .

Answer

1+6x+27x^2+108x^3+\cdots,\quad |x|<\tfrac13

Worked example

Surd: expand $\sqrt{1+x}$ up to $x^3$ , with validity

A square root is the power $n=\tfrac12$ , so the falling-factorial top runs $\tfrac12,\ -\tfrac12,\ -\tfrac32$ : $1+\tfrac12 x+\tfrac{\frac12(-\frac12)}{2!}x^2+\tfrac{\frac12(-\frac12)(-\frac32)}{3!}x^3$ .
Tidy the fractions: $\tfrac12,\ -\tfrac18,\ \tfrac1{16}$ .

Answer

1+\tfrac12 x-\tfrac18 x^2+\tfrac{1}{16}x^3+\cdots,\quad |x|<1

Worked example

Coefficient of $x^3$ in $\dfrac{x-5}{1+2x}$

Write as $(x-5)(1+2x)^{-1}$ and expand $(1+2x)^{-1}=1-2x+4x^2-8x^3+\cdots$ .
Collect the $x^3$ terms from the product: $x\cdot4x^2+(-5)(-8x^3)=4x^3+40x^3$ .

Answer

\text{coefficient}=44

Where the marks go

Common mistake

Always write the validity, and read it off the inside:

(1+2x)^n

is valid for

|x|<\tfrac12

, not

|x|<1

. People write

|x|<1

out of habit. Don't.

Common mistake

One wrong sign in the falling-factorial

n(n-1)(n-2)\cdots

flips every term after it, so check each coefficient. When

n

is negative the signs should switch plus, minus, plus, minus. If one does not, that is your warning.

Now you try

Expand $\sqrt[3]{1+6x}$ in ascending powers of $x$ up to the term in $x^3$ , simplify the coefficients, and give the validity. (9709/32 Oct/Nov 2020 Q2)

Your turn— tap to reveal the worked answer (9709/32 Oct/Nov 2020 Q2)

A cube root is the power $n=\tfrac13$ , so $\sqrt[3]{1+6x}=(1+6x)^{1/3}$ , with inside $6x$ . Falling-factorial top $\tfrac13,\ -\tfrac23,\ -\tfrac53$ :

$1+\tfrac13(6x)+\tfrac{\frac13(-\frac23)}{2!}(6x)^2+\tfrac{\frac13(-\frac23)(-\frac53)}{3!}(6x)^3$

Validity from the inside: $|6x|<1$ , so $|x|<\tfrac16$ .

Answer

1+2x-4x^2+\tfrac{40}{3}x^3+\cdots,\quad |x|<\tfrac16

PreviousPartial fractions NextBinomial expansion of (a + x)ⁿ

Binomial expansion for rational n

The binomial expansion no longer needs a whole-number power. It still works for any rational n, with two changes.
The series never stops. And it only works when $|x|<1$ .
In return you can now expand reciprocals (1 over something) and surds.

The general formula

$(1+x)^n=1+nx+\tfrac{n(n-1)}{2!}x^2+\tfrac{n(n-1)(n-2)}{3!}x^3+\cdots\quad(|x|<1)$

For each new term, multiply the last factor on top by the next number down from $n$ , then divide by the next factorial.
A square root is the power $n=\tfrac12$ , a cube root is $n=\tfrac13$ , and 1 over something is $n=-1$ .
If the inside is $kx$ , the range gets smaller: $|kx|<1$ , which means $|x|<\tfrac1{|k|}$ .

See the validity range

Drag $x$ across the line $x=1$ . Inside the band the blue running totals settle onto the gold true value of $(1+x)^n$ . Push $|x|$ to 1 or past it and they scatter.
That is the validity condition the mark scheme wants you to write down.

x = 0.50·true value 0.667·gap after 9 terms 0.001

The link to geometric series

Worked examples

Worked example

Expand $(1+x)^{-2}$ up to $x^3$ , with validity

With $n=-2$ the falling-factorial top runs $-2,\,-3,\,-4$ : $1+(-2)x+\tfrac{(-2)(-3)}{2}x^2+\tfrac{(-2)(-3)(-4)}{6}x^3$ .
Simplify each coefficient: $-2,\ \tfrac{6}{2}=3,\ \tfrac{-24}{6}=-4$ .

Answer

1-2x+3x^2-4x^3+\cdots,\quad |x|<1

Worked example

Textbook: expand $(1-3x)^{-2}$ up to $x^3$ , with validity

Use $n=-2$ and replace $x$ by $-3x$ : $1+(-2)(-3x)+\tfrac{(-2)(-3)}{2!}(-3x)^2+\tfrac{(-2)(-3)(-4)}{3!}(-3x)^3$ .
Each $(-3x)^k$ brings a power of $3$ : $6x,\ 3(9x^2),\ -4(-27x^3)$ , i.e. $6x,\ 27x^2,\ 108x^3$ .
Validity comes from the inside: $|-3x|<1$ , so $|x|<\tfrac13$ .

Answer

1+6x+27x^2+108x^3+\cdots,\quad |x|<\tfrac13

Worked example

Surd: expand $\sqrt{1+x}$ up to $x^3$ , with validity

A square root is the power $n=\tfrac12$ , so the falling-factorial top runs $\tfrac12,\ -\tfrac12,\ -\tfrac32$ : $1+\tfrac12 x+\tfrac{\frac12(-\frac12)}{2!}x^2+\tfrac{\frac12(-\frac12)(-\frac32)}{3!}x^3$ .
Tidy the fractions: $\tfrac12,\ -\tfrac18,\ \tfrac1{16}$ .

Answer

1+\tfrac12 x-\tfrac18 x^2+\tfrac{1}{16}x^3+\cdots,\quad |x|<1

Worked example

Coefficient of $x^3$ in $\dfrac{x-5}{1+2x}$

Write as $(x-5)(1+2x)^{-1}$ and expand $(1+2x)^{-1}=1-2x+4x^2-8x^3+\cdots$ .
Collect the $x^3$ terms from the product: $x\cdot4x^2+(-5)(-8x^3)=4x^3+40x^3$ .

Answer

\text{coefficient}=44

Where the marks go

Common mistake

Always write the validity, and read it off the inside:

(1+2x)^n

is valid for

|x|<\tfrac12

, not

|x|<1

. People write

|x|<1

out of habit. Don't.

Common mistake

One wrong sign in the falling-factorial

n(n-1)(n-2)\cdots

flips every term after it, so check each coefficient. When

n

is negative the signs should switch plus, minus, plus, minus. If one does not, that is your warning.

Now you try

Expand $\sqrt[3]{1+6x}$ in ascending powers of $x$ up to the term in $x^3$ , simplify the coefficients, and give the validity. (9709/32 Oct/Nov 2020 Q2)

Your turn— tap to reveal the worked answer (9709/32 Oct/Nov 2020 Q2)

A cube root is the power $n=\tfrac13$ , so $\sqrt[3]{1+6x}=(1+6x)^{1/3}$ , with inside $6x$ . Falling-factorial top $\tfrac13,\ -\tfrac23,\ -\tfrac53$ :

$1+\tfrac13(6x)+\tfrac{\frac13(-\frac23)}{2!}(6x)^2+\tfrac{\frac13(-\frac23)(-\frac53)}{3!}(6x)^3$

Validity from the inside: $|6x|<1$ , so $|x|<\tfrac16$ .

Answer

1+2x-4x^2+\tfrac{40}{3}x^3+\cdots,\quad |x|<\tfrac16