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General Binomial Expansion (Edexcel A Level Maths: Pure): Flashcards

Exam code: 9MA0

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FrontGeneral Binomial Expansion
Why does the expansion of $(1 + x)^{n}$ have infinitely many terms when $n$ is negative or fractional?

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Why does the expansion of $(1 + x)^{n}$ have infinitely many terms when $n$ is negative or fractional?
Each new term brings in the next factor of $n$ , $n - 1$ , $n - 2$ , and so on.
If $n$ is a positive integer these eventually reach $n - n = 0$ , and every term after that is zero, so the expansion stops.
A negative or fractional $n$ never hits zero, so the terms never run out.
Define the validity statement of a binomial expansion.
The range of values of $x$ for which an infinite binomial expansion converges, and so genuinely equals the expression it came from.
For $(1 + x)^{n}$ it is $| x | < 1$ , which is another way of writing $- 1 < x < 1$ .
Exam questions usually call this the range of validity.
Fill in the missing numerator:
$(1 + x)^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{______}{3!} x^{3} + \dots$
The completed expansion is:
$(1 + x)^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots$
Each term brings in one more factor, going down by 1 each time, and the denominator is the factorial of the power of $x$ .
True or False?
The expansion of $(1 + x)^{n}$ is only valid for $| x | < 1$ , whatever the value of $n$ .
False.
When $n$ is a positive integer the expansion is a finite sum, and it is valid for every value of $x$ .
The restriction $| x | < 1$ is only needed when the expansion is an infinite series, because an infinite series has to converge.
For practical purposes, why do you only need the first few terms of a general binomial expansion?
Because $| x | < 1$ , so raising $x$ to a high power gives a number very close to zero: $x^{r} \to 0$ as $r \to \infty$ .
The later terms are too small to make a noticeable difference, so a few terms give a good approximation, and it is rare to need to go beyond the term in $x^{3}$ .
The expansion only equals the original expression when every term is included.
In a general binomial expansion, what is the most common slip when working out $n - 1$ and $n - 2$ for a fractional or negative $n$ ?
Forgetting that the numbers become negative.
If $n = \frac{1}{2}$ then $n - 1 = - \frac{1}{2}$ and $n - 2 = - \frac{3}{2}$ .
If $n = - \frac{1}{3}$ then $n - 1 = - \frac{4}{3}$ .
How do you use the expansion of $(1 + x)^{n}$ to expand $(1 + b x)^{n}$ ?
Replace every $x$ in the expansion with $b x$ .
Keep $b x$ in brackets, so that, for example, $(b x)^{2} = b^{2} x^{2}$ , and check whether $b$ is negative: in ${(1 - \frac{2}{3} x)}^{n}$ , $b = - \frac{2}{3}$ .
What happens to the validity statement when you expand $(1 + b x)^{n}$ ?
Replace $x$ with $b x$ there too, giving $| b x | < 1$ .
For ${(1 - \frac{2}{3} x)}^{- 2}$ this is $|- \frac{2}{3} x| < 1$ , so $| x | < \frac{3}{2}$ .
The general binomial expansion is written for $(1 + x)^{n}$ .
What is the first thing you must do to expand $(a + b x)^{n}$ when $a \neq 1$ ?
Factorise $a$ out of the bracket, so the term inside becomes 1:
$(a + b x)^{n} = {[a (1 + \frac{b}{a} x)]}^{n} = a^{n} {(1 + \frac{b}{a} x)}^{n}$
The $a^{n}$ stays outside as a multiplier, and you expand the bracket.
Fill in the missing index:
$\frac{1}{\sqrt[3]{8 - 3 x}} = (8 - 3 x)^{______}$
The completed expression is:
$\frac{1}{\sqrt[3]{8 - 3 x}} = (8 - 3 x)^{- \frac{1}{3}}$
A root becomes a fractional power, and moving the bracket out of the denominator makes that power negative.
Why must ${(a + b x)}^{n}$ be rewritten in the form ${(1 + \frac{b}{a} x)}^{n}$ before the general binomial expansion can be used?
The expansion in the formula booklet is only given for $(1 + x)^{n}$ , with a 1 as the first term in the bracket.
Once the bracket is in that form you can read the expansion straight off the booklet and replace $x$ with $\frac{b}{a} x$ .
What is the range of validity of the expansion of $(a + b x)^{n}$ ?
$|\frac{b}{a} x| < 1$ , which rearranges to $| x | < |\frac{a}{b}|$ .
It comes from the bracket after factorising, not from the original expression. For $(8 - 3 x)^{- \frac{1}{3}} = 8^{- \frac{1}{3}} {(1 - \frac{3}{8} x)}^{- \frac{1}{3}}$ it gives $| x | < \frac{8}{3}$ .
True or False?
The expansions of $(3 + 2 x)^{- 4}$ and $(1 + 2 x)^{- 4}$ are valid for the same values of $x$ .
False.
The range of validity depends on $\frac{b}{a}$ , so changing $a$ changes it.
$(3 + 2 x)^{- 4} = 3^{- 4} {(1 + \frac{2}{3} x)}^{- 4}$ is valid for $| x | < \frac{3}{2}$ , while $(1 + 2 x)^{- 4}$ is valid for $| x | < \frac{1}{2}$ .
How do you expand an expression containing more than one binomial, such as $\frac{\sqrt{1 + x}}{3 + 2 x}$ ?
Break it into separate binomials, here $(1 + x)^{\frac{1}{2}}$ and $(3 + 2 x)^{- 1}$ .
Expand each one individually, then multiply the expansions together and collect like terms.
True or False?
To multiply two binomial expansions together up to the term in $x^{2}$ , you must multiply out every pair of terms.
False.
Any product whose powers add to more than 2 can be ignored, since it only affects terms you are not keeping.
So you only need the pairs that give $x^{0}$ , $x^{1}$ and $x^{2}$ , which saves a great deal of work.
When expanding an expression that contains more than one binomial, how far must each one be expanded if the final answer is needed up to the term in $x^{3}$ ?
As far as the term in $x^{3}$ in each expansion.
A term in $x^{3}$ in the final answer can come from $1 \times x^{3}$ as well as from $x \times x^{2}$ , so stopping any earlier would lose part of it.
Expanding $\frac{\sqrt{1 + x}}{3 + 2 x}$ uses $(1 + x)^{\frac{1}{2}}$ , valid for $| x | < 1$ , and $(3 + 2 x)^{- 1}$ , valid for $| x | < \frac{3}{2}$ .
The whole expansion is valid for $______$
The whole expansion is valid for $| x | < 1$ .
Both expansions have to be valid at the same time, so the overall range of validity is the intersection of the two: the smaller boundary wins.
What lets you apply the general binomial expansion to a rational function such as $\frac{9 x + 10}{(x + 4) (3 x - 1)}$ ?
Splitting it into partial fractions first:
$\frac{9 x + 10}{(x + 4) (3 x - 1)} = \frac{2}{x + 4} + \frac{3}{3 x - 1}$
Each partial fraction can then be written as a negative power, $2 (x + 4)^{- 1} + 3 (3 x - 1)^{- 1}$ , and expanded.
How do you prepare $3 (3 x - 1)^{- 1}$ for a binomial expansion, when the constant term is $- 1$ ?
Factorise the $- 1$ out of the bracket so the constant term becomes $+ 1$ :
$3 (- 1 + 3 x)^{- 1} = 3 {[(- 1) (1 - 3 x)]}^{- 1} = - 3 (1 - 3 x)^{- 1}$
The $- 1$ leaves the bracket raised to the power $n$ , here $(- 1)^{- 1} = - 1$ .
How do you use a binomial expansion to approximate a numerical value?
Compare the number you want with the expression that was expanded, solve for $x$ , then substitute that $x$ into the expansion.
To approximate $\sqrt[4]{85}$ from $\sqrt[4]{81 - 9 x}$ , solve $81 - 9 x = 85$ to get $x = - \frac{4}{9}$ , and put that into the expansion.
What makes a binomial approximation more accurate?
Using more terms of the expansion. Each extra term brings the value closer to the true one.
Terms up to $x^{2}$ or $x^{3}$ are usually accurate enough.
Before using a value of $x$ in a binomial approximation, what must you check about it?
That it lies inside the range of validity of the expansion.
$\sqrt[4]{81 - 9 x}$ is only valid for $| x | < 9$ , so $\sqrt[4]{171}$ , which needs $x = - 10$ , cannot be approximated from it.
Exam questions often hide a validity check inside an approximation question.
True or False?
If the value of $x$ you need lies outside the range of validity of a binomial expansion, using more terms will still give a good approximation.
False.
Outside the range of validity the series does not converge, so extra terms do not settle towards the true value, they make things worse.
The expansion cannot be used for that value at all.
Fill in the missing values so a binomial expansion can be used to approximate $\sqrt{710}$ :
$\sqrt{710} = \sqrt{______\times 7.1} =______\sqrt{7.1}$
The completed working is:
$\sqrt{710} = \sqrt{100 \times 7.1} = 10 \sqrt{7.1}$
Taking out a perfect square leaves a much smaller number, which an expansion can reach.
Two expansions can both approximate the same value, one needing $x = 0.04$ and the other $x = 0.8$ .
Which gives the better approximation?
The one using $x = 0.04$ .
The terms left out are powers of $x$ , and those shrink far faster when $x$ is small. A value of $x$ near the edge of the range of validity gives a poor approximation even though the expansion is still valid.

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