What is 4.1: Binomial expansion?

For a positive integer n, (a+bx)^n=/ra^n-r(bx)^r, and the same coefficients give binomial probabilities n/rp^r(1-p)^n-r. For a rational or negative n use (1+x)^n=1+nx+n(n-1)/2!x^2+n(n-1)(n-2)/3!x^3+, an infinite series valid only for |x|<1.

How many marks are 4.1 questions worth?

4.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 4.1 appeared in?

We have not yet indexed a verified past-paper appearance for 4.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

4.1

Understand and use the binomial expansion of (a + bx)ⁿ for positive integer n; the notations n! and nCr; link to binomial probabilities; extend to any rational n, including for approximation, valid for |bx/a| < 1.

Q: How many marks are 4.1 questions worth?

4.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Q: Which past exam papers has 4.1 appeared in?

We have not yet indexed a verified past-paper appearance for 4.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

Draft — not yet indexed

Binomial expansion

Worked answers and methods for 4.1 on Edexcel A-level Maths 9MA0.

Definition

Specification 4.1: understand and use the binomial expansion of $(a+bx)^n$ for positive integer $n$ ; the notations $n!$ and $\binom{n}{r}$ ; link to binomial probabilities; extend to any rational $n$ , including its use for approximation; be aware that the expansion is valid for $\left|\dfrac{bx}{a}\right|<1$ .

Explanation

For a positive integer $n$ , $(a+bx)^n=\sum\binom{n}{r}a^{n-r}(bx)^r$ , and the same coefficients give binomial probabilities $\binom{n}{r}p^r(1-p)^{n-r}$ . For a rational or negative $n$ use $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots$ , an infinite series valid only for $|x|<1$ .
Replace $x$ by the whole of $bx$ in brackets, including its sign and coefficient, so $(1-9x)^{1/2}$ has second term $\tfrac12(-9x)$ and third $\dfrac{\frac12\cdot(-\frac12)}{2}(-9x)^2$ .
When the first term is not $1$ , take it out first: $\sqrt{4+5x}=2(1+\tfrac54x)^{1/2}$ , and the validity becomes $\left|\tfrac54x\right|<1$ .
State the range of validity, $|x|<\dfrac{1}{9}$ or $|9x|<1$ , and check a substituted value against it before using the series for an approximation; a value outside the range gives no approximation at all.
Simplify each coefficient fully.

Worked example

Find the first three terms, in ascending powers of

x

, of the binomial expansion of

(1-2x)^{-\frac12}

, giving each coefficient in its simplest form. State the range of values of

x

for which the expansion is valid.

(4)

(Total for Question 1 is 4 marks)

Working

1.Use $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\cdots$ with $n=-\dfrac12$ and $u=-2x$ .
2. $1+\left(-\dfrac12\right)(-2x)+\dfrac{(-\frac12)(-\frac32)}{2}(-2x)^2=1+x+\dfrac{3}{8}\times4x^2$ .
3.So the expansion is $1+x+\dfrac32x^2+\cdots$ , valid for $|-2x|<1$ , that is $|x|<\dfrac12$ .

Answer: $1+x+\dfrac32x^2$ ; valid for $|x|<\dfrac12$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Attempts the binomial expansion with index $-\frac12$ and the correct structure for the $x^2$ term, eg $\dfrac{(-\frac12)(-\frac32)}{2!}(\pm2x)^2$	M1
	Correct unsimplified expansion with brackets, $1+(-\frac12)(-2x)+\ldots$	A1
	$1+x+\dfrac32x^2$	A1
	$\|x\|<\dfrac12$ or $-\dfrac12<x<\dfrac12$	B1
Notes
A1: cao
B1: oe

(4 marks)

Common mistakes

Don't expand $(a+bx)^n$ without first factoring out $a$ , so the generalised binomial coefficients are applied to the wrong expression.
Don't use the finite binomial formula for a negative or fractional power and omit the convergence condition.

Mark-scheme wording to use

“M1: For an attempt at the binomial expansion with n = 1/2 and obtains the correct structure for term 3 or term 4. … A1: Correct unsimplified expression as shown but the bracketing must be correct unless any missing brackets are implied by subsequent work. … OR allow this mark for at least 2 correct simplified terms from −(9/2)x, −(81/8)x² and −(729/16)x³ … A1: 1 − (9/2)x − (81/8)x² − (729/16)x³ or simplified equivalent.” — Pearson 9MA0, June 2024 Paper 1, Q2(a)
“B1: Expansion is valid for |x| < 1/9 or |x| ≤ 1/9 and x = −2/9 is outside this range. … Do not accept vague statements such as "it is too big", "it is outside the range" without any mention of what the range is.” — Pearson 9MA0, June 2024 Paper 1, Q2(b)
“B1 for extracting the factor 4^(1/2), equivalently writing 2(1 + 5x/4)^(1/2). M1 for applying the binomial expansion to at least two non-constant terms. A1ft for a correct consistent expansion through x^2. A1 for the fully correct result k = -25/64” — Pearson 9MA0, Specimen Paper 1, Q2(a)

What the examiner wants

Brackets round the whole term and a stated range. The June 2024 Paper 1 report says the binomial structure was generally very good but inconsistency with brackets and not simplifying cost accuracy marks, the most common error being $9$ instead of $-9$ ; on validity, many could not state $|x|<\frac19$ and misread the modulus, writing $x>\frac19$ .

Exam tip

For a rational-power expansion, factor the constant first and state the validity condition from the transformed bracket.

Worked practice

Tier 1 · Easy

Find the coefficient of

x^2

in the expansion of

(1+4x)^5

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$160$	2
Notes
The $x^2$ term is obtained with $r=2$ : $\binom{5}{2}(4x)^2=10\times16x^2=160x^2$ . Hence the coefficient is $160$ .

(2 marks)

Tier 2 · Standard

Find the coefficient of

x^3

in the expansion of

(2-x)^7

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$-560$	3
Notes
The $x^3$ term is obtained by choosing three factors of $-x$ : $\binom{7}{3}2^{7-3}(-x)^3=35\times16\times(-x^3)=-560x^3$ . Hence the coefficient is $-560$ .

(3 marks)

Tier 3 · Hard

Use the first three terms of a binomial expansion of

(64+x)^{1/3}

to estimate

\sqrt[3]{65}

. Give the estimate to

5

decimal places and justify that the expansion is valid at the value of

x

used.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\sqrt[3]{65}\approx4.02072$ Valid because $\|1/64\|<1$	5
Notes
Write $(64+x)^{1/3}=4(1+x/64)^{1/3}$ . Using $n=1/3$ , the first three terms give $4[1+x/192-x^2/36864]=4+x/48-x^2/9216$ . Set $x=1$ to estimate $\sqrt[3]{65}$ : $4+1/48-1/9216=4.0207248\ldots$ , so the estimate is $4.02072$ . The expansion requires $\|x/64\|<1$ , which holds when $x=1$ .

(5 marks)

Tier 1 · Easy

Expand

(3-2x)^4

up to and including the term in

x^2

. Write the result in ascending powers of

x

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$81-216x+216x^2$	3
Notes
The required terms are $3^4+\binom41 3^3(-2x)+\binom42 3^2(-2x)^2$ . These simplify to $81-216x+216x^2$ .

(3 marks)

Tier 2 · Standard

Expand

(1+3x)^{-2}

up to and including the term in

x^3

, giving the terms in ascending powers of

x

. State the range of

x

-values for which the expansion is valid.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$1-6x+27x^2-108x^3$ $\|x\|<\dfrac13$	5
Notes
Using $(1+u)^{-2}=1-2u+3u^2-4u^3+\cdots$ with $u=3x$ gives $1-6x+27x^2-108x^3+\cdots$ . Validity requires $\|3x\|<1$ , so $\|x\|<1/3$ .

(5 marks)

Tier 3 · Hard

In the expansion of

(1+px)(2-x)^6

, the term in

x^2

vanishes, where

p

is a constant. Find

p

and hence determine the coefficient of

x^3

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$p=\dfrac54$ Coefficient of $x^3$ is $140$	5
Notes
In $(2-x)^6$ , the coefficients of $x$ , $x^2$ and $x^3$ are $-192$ , $240$ and $-160$ respectively. The $x^2$ coefficient in the product is therefore $240-192p$ . Setting this equal to zero gives $p=5/4$ . The $x^3$ coefficient is then $-160+p(240)=-160+300=140$ .

(5 marks)

Tier 2 · Standard

Find the expansion of

\dfrac{(1+2x)^{1/2}}{1-x}

through the term in

x^2

, arranging the terms by ascending powers of

x

. State the set of

x

-values for which both component expansions are valid.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$1+2x+\dfrac32x^2$ $-\dfrac12<x<\dfrac12$	5
Notes
Use $(1+2x)^{1/2}=1+x-\dfrac12x^2+\cdots$ and $(1-x)^{-1}=1+x+x^2+\cdots$ . Multiplying and collecting terms through $x^2$ gives $1+2x+\dfrac32x^2$ . The first expansion requires $\|2x\|<1$ and the second requires $\|x\|<1$ , so both are valid when $-1/2<x<1/2$ .

(5 marks)

Tier 3 · Hard

Let

X

have the binomial distribution

B(8,p)

, where

0<p<1

. In the expansion of

(1-p+px)^8

, the coefficient of

x^4

10

times the coefficient of

x^2

. Find

p

. Hence find the exact value of

P(X\geq7)

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$p=\dfrac23$ $P(X\geq7)=\dfrac{1280}{6561}$	5
Notes
The coefficients of $x^4$ and $x^2$ are $\binom84p^4(1-p)^4$ and $\binom82p^2(1-p)^6$ . Their ratio is $\dfrac52[p/(1-p)]^2$ . Equating this to $10$ gives $[p/(1-p)]^2=4$ . Since $0<p<1$ , $p/(1-p)>0$ , so $p/(1-p)=2$ and $p=2/3$ . Hence $P(X\geq7)=\binom87(2/3)^7(1/3)+(2/3)^8=1280/6561$ .

(5 marks)

Tier 3 · Hard

The expansion of

(1+kx)^p

begins

1+6x+3x^2+\cdots

, where

k>0

and

p

is rational. Find

p

and

k

. Hence find the coefficient of

x^3

and state the set of

x

-values for which the expansion is valid.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$p=\dfrac65$ , $k=5$ Coefficient of $x^3$ is $-4$ $\|x\|<\dfrac15$	6
Notes
The coefficients of $x$ and $x^2$ give $pk=6$ and $p(p-1)k^2/2=3$ . Since $k=6/p$ , substitution in the second equation gives $18(p-1)/p=3$ , so $p=6/5$ and $k=5$ . The coefficient of $x^3$ is $[p(p-1)(p-2)/6]k^3=[(6/5)(1/5)(-4/5)/6]125=-4$ . The generalised binomial expansion requires $\|kx\|<1$ , hence $\|x\|<1/5$ .

(6 marks)

Q10

Tier 3 · Hard

10.

Use the first three terms of the binomial expansion of

(1+u)^{1/2}

to obtain a rational estimate for

\sqrt{63}

. State why the chosen value of

u

meets the validity condition. By comparing the square of the estimate with

63

, determine whether it is an overestimate or an underestimate.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\sqrt{63}\approx\dfrac{32511}{4096}$ Valid because $\|u\|=\dfrac{1}{64}<1$ It is an overestimate	6
Notes
Write $\sqrt{63}=8(1-1/64)^{1/2}$ , so $u=-1/64$ . Using $(1+u)^{1/2}=1+u/2-u^2/8+\cdots$ gives $8[1-1/128-1/32768]=32511/4096$ . The expansion is valid because $\|u\|=1/64<1$ . Also $(32511/4096)^2-63=513/16777216>0$ . Since the estimate is positive, its square being greater than $63$ shows that it is an overestimate of $\sqrt{63}$ .

(6 marks)

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