International A Level (IAL)MathsEdexcelPure 4FlashcardsBinomial ExpansionGeneral Binomial Expansion

General Binomial Expansion (Edexcel International A Level (IAL) Maths: Pure 4): Flashcards

Exam code: YMA01

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FrontGeneral Binomial Expansion
Define the general binomial expansion.

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Define the general binomial expansion.
The general binomial expansion is the expansion of ${(1 + x)}^{n}$ for any real $n$ , not merely for positive integers.
In practice that means negative and fractional powers, which the ordinary binomial expansion cannot handle.
Complete the general binomial expansion:
${(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{______}{3!} x^{3} + \dots$
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 numerator picks up one more factor than the last, each one lower by $1$ , and the factorial underneath keeps pace.
True or False?
The general binomial expansion of ${(1 + x)}^{n}$ always has infinitely many terms.
False.
When $n$ is a positive integer one of the numerator factors becomes zero, so the expansion stops there and is exact.
For any other real value of $n$ it does run on for ever.
For which values of $x$ is the expansion of ${(1 + x)}^{n}$ valid?
Only for $|x| < 1$ , which is another way of writing $- 1 < x < 1$ .
This is called the validity statement, and outside that range the infinite series does not add up to the original expression at all.
Why does the expansion only work when $|x| < 1$ ?
Because a number smaller than $1$ in size gets smaller still when raised to a higher power.
That is what makes the terms shrink towards zero, so the series converges rather than growing without limit.
How do you expand ${(1 + b x)}^{n}$ ?
Replace every $x$ in the standard expansion by $b x$ , watching the sign carefully if $b$ is negative.
The validity condition travels with it and becomes $|b x| < 1$ .
Why is the general expansion written for ${(1 + x)}^{n}$ rather than ${(a + b)}^{n}$ ?
Because setting $a = 1$ makes every power of $a$ equal to $1$ , so all those factors disappear.
What is left is short enough to write down and work with, which the general two-letter form is not.
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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