Evaluate 72: 72=(7!)/(2! 5!)=21 Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r State the final answer: 21 Answer: 21

A-Level Binomial expansion Practice Questions

Free A-Level Binomial expansion practice questions with full step-by-step worked solutions. Covers binomial coefficient, nCr, special cases, Pascal's triangle. Practise exam-style problems and check your method.

binomial coefficientnCrspecial casesPascal's trianglesum of coefficientsexpansion

A-Level70 questionsStep-by-step solutions

Question 1

2 markseasy

Evaluate

\binom{7}{2}

Show worked solution

Worked solution

Evaluate \binom{7}{2}
$\binom{7}{2}=\frac{7!}{2!\,5!}=21$
The coefficient \binom{7}{2} is worked out with the nCr formula, which counts the ways of choosing 2 things from 7. You can also read it straight from row 7 of Pascal's triangle.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
State the final answer
$21$
This value is the answer. A quick sanity check: coefficients like this are always whole numbers.

Answer

21

Question 2

3 markseasy

Expand

(1 - 2x)^{3}

, simplifying each term.

Show worked solution

Worked solution

Write down the binomial theorem
$(a+bx)^n=\sum_{r=0}^{3}\binom{3}{r}a^{3-r}(bx)^r$
The binomial theorem builds the expansion one term at a time. Here a and b are the two parts of the bracket and n is the power.
Identify a, b and n
$a=1,\quad b=-2,\quad n=3$
We match our bracket to the general form so we know exactly what to substitute. Getting these right at the start prevents mistakes later.
Write down row 3 of Pascal's triangle
$1,\ 3,\ 3,\ 1$
These are the binomial coefficients \binom{3}{0} up to \binom{3}{3}. Reading them from Pascal's triangle is quicker than using factorials each time.
Form each term
$\binom{3}{0}(1)^{3}(-2x)^{0} + \binom{3}{1}(1)^{2}(-2x)^{1} + \binom{3}{2}(1)^{1}(-2x)^{2} + \binom{3}{3}(1)^{0}(-2x)^{3}$
Each term multiplies a coefficient by a falling power of a and a rising power of the x-part. Notice the powers always add up to n.
Simplify every term
$1 - 6x + 12x^{2} - 8x^{3}$
Now we work out each bracket and multiply the numbers. Collecting them in ascending powers of x gives the tidy expansion.

Answer

1 - 6x + 12x^{2} - 8x^{3}

Question 3

4 marksintermediate

Using the first three terms of the expansion of

(1+x)^{10}

, estimate the value of

1.01^{10}

Show worked solution

Worked solution

Write the first three terms of (1+x)^{10}
$(1+x)^{10}\approx 1+10x+45x^2$
For an approximation we only keep the low powers of x, because higher powers of a small number are tiny. Here \binom{10}{1}=10 and \binom{10}{2}=45.
Choose x to match the number
$1+x=1.01\Rightarrow x=0.01$
We pick x so that 1+x equals the number we want a power of. Since 1.01=1+0.01, we use x=0.01.
Substitute x=0.01
$1+10(0.01)+45(0.01)^2$
Putting x=0.01 into the three terms turns the algebra into arithmetic. Because x is small, the x^2 term is already very small.
Work out each term
$1+0.1+45(0.0001)=1+0.1+0.0045$
We evaluate the pieces separately to avoid slips. The last term 0.0045 is small, showing why three terms is usually enough.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Add them up
$1.01^{10}\approx 1.1045$
Adding the three values gives the estimate. This is very close to the true value 1.10462..., so the approximation is good.

Answer

1.1045

Question 4

5 markshard

Use the first three terms of the expansion of

(1+x)^{15}

to estimate

1.002^{15}

5

decimal places.

Show worked solution

Worked solution

First three terms of (1+x)^{15}
$1+15x+105x^2$
We keep terms up to x^2. Here \binom{15}{1}=15 and \binom{15}{2}=105.
Choose x
$1+x=1.002\Rightarrow x=0.002$
Matching 1+x to 1.002 gives x=0.002, which is very small. Small x makes the approximation accurate.
Substitute
$1+15(0.002)+105(0.002)^2$
Substituting turns the expansion into a numerical calculation. The x^2 term will be very small.
Evaluate the terms
$1+0.03+0.00042$
We work out each piece: 15x0.002=0.03 and 105x0.000004=0.00042. Both are easy to add.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Add up
$1.002^{15}\approx 1.03042$
Adding gives 1.03042, matching the true value to five decimal places.

Answer

1.03042

Question 5

8 markschallenging

Explain why the expansion of

(a+bx)^n

(for a positive integer

n

) has exactly

n+1

terms.

Show worked solution

Worked solution

Recall the general term
$\binom{n}{r}a^{n-r}b^r,\quad r=0,1,\dots,n$
Each term of (a+b)^n is set by a value of r. So the possible values of r decide how many terms there are.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Recall Pascal's triangle relation
$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$
Each entry in Pascal's triangle is the sum of the two above it. This is where the coefficients come from without any factorials.
Note the end coefficients
$\binom{n}{0}=1,\qquad \binom{n}{n}=1$
The very first and very last coefficients are always 1. That gives you two terms of the answer for free.
Be careful with the signs
$(-1)^r>0 \text{ when } r \text{ is even, } <0 \text{ when } r \text{ is odd}$
When the bracket contains a minus sign the terms alternate in sign. Track this carefully so no negative is lost.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Count the values of r
$r\in\{0,1,2,\dots,n\}\Rightarrow n+1 \text{ values}$
The counter r runs from 0 up to n inclusive. Counting these values, there are n+1 of them, so there are n+1 terms.

Answer

Because r takes the n+1 values 0,1,...,n.

Unlock 65 more Binomial expansion questions

Create a free account to work through every A-Level Binomial expansion question with instant step-by-step worked solutions, progress tracking and interactive lessons.

Full worked solutions for every question
Interactive lessons and instant feedback
Track your mastery across every topic

Create a Free Account

No card required · Free forever

More Binomial expansion practice

Hard Questions Worked Solutions

A-Level Binomial expansion Practice Questions

Worked solution

Worked solution

Worked solution

Worked solution

Worked solution

Unlock 65 more Binomial expansion questions

More Binomial expansion practice

Related Pure Maths topics