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Understanding the Binomial Theorem for WJEC AS-level Pure Mathematics

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Megan

06/12/2025

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WJEC AS-level Pure Mathematics Binomial theorem

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6 Dec 2025

10 pages

Understanding the Binomial Theorem for WJEC AS-level Pure Mathematics

user profile picture

Megan

@megan_0306

The binomial theoremis one of the most powerful tools... Show more

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Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Building Up to the Binomial Theorem

Ever wondered if there's a shortcut to expanding 1+x1+x⁴ without multiplying brackets four times? There absolutely is! Let's start by looking at the pattern that emerges when we expand simple binomial expressions.

When you expand 1+x1+x², you get 1+2x+x². For 1+x1+x³, the result is 1+3x+3x²+x³. Notice how the coefficients follow a specific pattern - this isn't coincidence, it's the foundation of the binomial theorem.

The key insight is that these coefficients come from Pascal's triangle. Each row gives you the coefficients for the next power, making expansions much quicker than traditional multiplication.

Quick Tip: Always check your expansions by substituting x=1 - both sides should give you the same result!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Pascal's Triangle and Advanced Expansions

Pascal's triangle is your best mate for binomial expansions. Each number is the sum of the two numbers above it, and each row gives you the coefficients for 1+x1+xⁿ where n is the row number.

The real magic happens when you need to expand something like 1+2x1+2x³ or 3+x3+x⁴. You use the same coefficients from Pascal's triangle, but you need to be careful with the powers. For 1+2x1+2x³, you substitute 2x wherever you see x in the basic expansion.

For expressions like 3+x3+x⁴, you can rewrite it as 3⁴1+x/31+x/3⁴ or use the fact that each term involves powers of both 3 and x. The coefficient pattern stays the same - it's just the arithmetic that gets trickier.

Remember: The coefficients from Pascal's triangle never change - only how you apply the powers of your variables changes.

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

The General Formula and Factorials

The general binomial theorem states that a+ba+bⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... where ⁿCᵣ represents "n choose r". This notation might look intimidating, but it's just a systematic way to find those Pascal's triangle numbers.

Factorials are crucial here - they're written as n! and mean n×n1n-1×n2n-2×...×1. So 5! = 5×4×3×2×1 = 120. The formula ⁿCᵣ = n!/r!(nr)!r!(n-r)! tells you exactly how many ways you can choose r objects from n objects.

Your calculator can handle factorials and combinations, but understanding the pattern helps you spot shortcuts. For instance, ⁵C₂ = (5×4)/(2×1) = 10, which is much quicker than calculating the full factorials.

Pro Tip: Learn to cancel common factors in factorial fractions - it'll save you loads of time in exams!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Applying the General Formula

Now you can tackle any binomial expansion systematically. For 1+x1+x¹⁵, you don't need to expand the whole thing - just find the terms you need using the general formula.

The general term ther+1termthe r+1 term is ⁿCᵣaⁿ⁻ʳbʳ. This formula lets you find specific terms without expanding everything. For the first four terms of 1+x1+x¹⁵, you calculate ¹⁵C₀, ¹⁵C₁, ¹⁵C₂, and ¹⁵C₃.

More complex expressions like 2xx22x-x²⁵ follow the same pattern, but you need to be extra careful with negative signs and powers. Each term alternates sign because of the x2-x² part, and the powers of x build up from both parts of the binomial.

Watch Out: Keep track of negative signs carefully - they follow a pattern but it's easy to make mistakes when you're rushing!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Finding Unknown Coefficients

Sometimes you'll be given information about coefficients and asked to find unknown values. These problems test your understanding of the binomial expansion formula and your algebra skills.

For example, if the coefficient of x⁴ is 12 times the coefficient of x³ in a+2xa+2x⁴, you set up an equation using the general formula. The coefficient of x³ is ⁴C₃×a×(2x)³ = 32a, and the coefficient of x⁴ would be from the next term.

Setting up the equation 32a × 12 = coefficient of x⁴ gives you 384a. But wait - there's no x⁴ term in a+2xa+2x⁴! This means you need to reconsider the problem setup and check your working carefully.

Strategy: Always write out the first few terms explicitly before setting up your equations - it prevents costly errors!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Complex Coefficient Problems

When dealing with more complex coefficient relationships, your algebraic manipulation skills become crucial. These problems often involve setting up equations where coefficients are related by specific ratios or differences.

Consider when the coefficient of x is 23040 smaller than the coefficient of x² in 2+ax2+ax⁹. You expand the first few terms, identify the relevant coefficients, and set up your equation. The algebra can get messy, but the method stays the same.

Sometimes you'll end up with quadratic equations in your unknown. Use the quadratic formula when factoring isn't obvious, and always check that your answers make sense in the original context.

Don't Panic: These problems look scary but they're just systematic applications of the binomial formula plus some algebra!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Working with Unknown Indices

Unknown indices problems give you information about coefficients and ask you to find the power n. These require you to use the relationship between different terms in the expansion.

If the 3rd and 5th terms have equal coefficients in 1+x1+xⁿ, you set ⁿC₂ = ⁿC₄. Using the factorial formula and simplifying gives you a quadratic equation in n. Remember to check that your answer satisfies any given conditions.

The key insight is that ⁿCᵣ = ⁿCₙ₋ᵣ, which explains why some coefficients can be equal. This symmetry in Pascal's triangle is often the foundation for these problems.

Remember: Pascal's triangle is symmetric - this symmetry is often the key to solving unknown index problems!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

More Practice with Unknown Indices

Let's solidify your understanding with another approach. When the coefficient of x² in 1+x1+xⁿ equals 55, you use ⁿC₂ = 55. This gives you nn1n-1/2 = 55, leading to n²-n-110 = 0.

For expressions like 1+2x1+2xⁿ where the coefficient of x² is 40, remember that the coefficient involves both the combination and the power of 2. So ⁿC₂ × 2² = 40, giving you ⁿC₂ = 10.

These problems follow the same pattern: identify the relevant term, extract the coefficient using the general formula, set up your equation, and solve. The algebra might vary, but the method is consistent.

Success Strategy: Practice identifying which combination formula you need - that's usually the trickiest part!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Coefficient Problems with Modified Expressions

When working with expressions like 1+2x1+2xⁿ, don't forget that the coefficient includes powers of the constant. For the x² term, you need ⁿC₂ × (2x)² = ⁿC₂ × 4x².

If this coefficient equals 40, then ⁿC₂ × 4 = 40, so ⁿC₂ = 10. This gives you nn1n-1/2 = 10, leading to n²-n-20 = 0. Factoring gives you n5n-5n+4n+4 = 0.

Since n must be positive, n = 5 is your answer. Always check that your solution makes sense in the context - negative powers or impossible values should make you reconsider your working.

Final Check: Substitute your answer back into the original expansion to verify your coefficient calculation!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3


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This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

Samantha Klich

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Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

Anna

iOS user

Best app on earth! no words because it’s too good

Thomas R

iOS user

Just amazing. Let's me revise 10x better, this app is a quick 10/10. I highly recommend it to anyone. I can watch and search for notes. I can save them in the subject folder. I can revise it any time when I come back. If you haven't tried this app, you're really missing out.

Basil

Android user

This app has made me feel so much more confident in my exam prep, not only through boosting my own self confidence through the features that allow you to connect with others and feel less alone, but also through the way the app itself is centred around making you feel better. It is easy to navigate, fun to use, and helpful to anyone struggling in absolutely any way.

David K

iOS user

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Android user

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Greenlight Bonnie

Android user

very reliable app to help and grow your ideas of Maths, English and other related topics in your works. please use this app if your struggling in areas, this app is key for that. wish I'd of done a review before. and it's also free so don't worry about that.

Rohan U

Android user

I know a lot of apps use fake accounts to boost their reviews but this app deserves it all. Originally I was getting 4 in my English exams and this time I got a grade 7. I didn’t even know about this app three days until the exam and it has helped A LOT. Please actually trust me and use it as I’m sure you too will see developments.

Xander S

iOS user

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Maths

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6 Dec 2025

10 pages

Understanding the Binomial Theorem for WJEC AS-level Pure Mathematics

user profile picture

Megan

@megan_0306

The binomial theoremis one of the most powerful tools in A-level maths, letting you expand expressions like (a+b)ⁿ without having to multiply everything out by hand. Once you grasp the pattern and learn to use Pascal's triangle or the... Show more

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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Building Up to the Binomial Theorem

Ever wondered if there's a shortcut to expanding 1+x1+x⁴ without multiplying brackets four times? There absolutely is! Let's start by looking at the pattern that emerges when we expand simple binomial expressions.

When you expand 1+x1+x², you get 1+2x+x². For 1+x1+x³, the result is 1+3x+3x²+x³. Notice how the coefficients follow a specific pattern - this isn't coincidence, it's the foundation of the binomial theorem.

The key insight is that these coefficients come from Pascal's triangle. Each row gives you the coefficients for the next power, making expansions much quicker than traditional multiplication.

Quick Tip: Always check your expansions by substituting x=1 - both sides should give you the same result!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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Access to all documents

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Join milions of students

By signing up you accept Terms of Service and Privacy Policy

Pascal's Triangle and Advanced Expansions

Pascal's triangle is your best mate for binomial expansions. Each number is the sum of the two numbers above it, and each row gives you the coefficients for 1+x1+xⁿ where n is the row number.

The real magic happens when you need to expand something like 1+2x1+2x³ or 3+x3+x⁴. You use the same coefficients from Pascal's triangle, but you need to be careful with the powers. For 1+2x1+2x³, you substitute 2x wherever you see x in the basic expansion.

For expressions like 3+x3+x⁴, you can rewrite it as 3⁴1+x/31+x/3⁴ or use the fact that each term involves powers of both 3 and x. The coefficient pattern stays the same - it's just the arithmetic that gets trickier.

Remember: The coefficients from Pascal's triangle never change - only how you apply the powers of your variables changes.

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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The General Formula and Factorials

The general binomial theorem states that a+ba+bⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... where ⁿCᵣ represents "n choose r". This notation might look intimidating, but it's just a systematic way to find those Pascal's triangle numbers.

Factorials are crucial here - they're written as n! and mean n×n1n-1×n2n-2×...×1. So 5! = 5×4×3×2×1 = 120. The formula ⁿCᵣ = n!/r!(nr)!r!(n-r)! tells you exactly how many ways you can choose r objects from n objects.

Your calculator can handle factorials and combinations, but understanding the pattern helps you spot shortcuts. For instance, ⁵C₂ = (5×4)/(2×1) = 10, which is much quicker than calculating the full factorials.

Pro Tip: Learn to cancel common factors in factorial fractions - it'll save you loads of time in exams!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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Applying the General Formula

Now you can tackle any binomial expansion systematically. For 1+x1+x¹⁵, you don't need to expand the whole thing - just find the terms you need using the general formula.

The general term ther+1termthe r+1 term is ⁿCᵣaⁿ⁻ʳbʳ. This formula lets you find specific terms without expanding everything. For the first four terms of 1+x1+x¹⁵, you calculate ¹⁵C₀, ¹⁵C₁, ¹⁵C₂, and ¹⁵C₃.

More complex expressions like 2xx22x-x²⁵ follow the same pattern, but you need to be extra careful with negative signs and powers. Each term alternates sign because of the x2-x² part, and the powers of x build up from both parts of the binomial.

Watch Out: Keep track of negative signs carefully - they follow a pattern but it's easy to make mistakes when you're rushing!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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Finding Unknown Coefficients

Sometimes you'll be given information about coefficients and asked to find unknown values. These problems test your understanding of the binomial expansion formula and your algebra skills.

For example, if the coefficient of x⁴ is 12 times the coefficient of x³ in a+2xa+2x⁴, you set up an equation using the general formula. The coefficient of x³ is ⁴C₃×a×(2x)³ = 32a, and the coefficient of x⁴ would be from the next term.

Setting up the equation 32a × 12 = coefficient of x⁴ gives you 384a. But wait - there's no x⁴ term in a+2xa+2x⁴! This means you need to reconsider the problem setup and check your working carefully.

Strategy: Always write out the first few terms explicitly before setting up your equations - it prevents costly errors!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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Access to all documents

Improve your grades

Join milions of students

By signing up you accept Terms of Service and Privacy Policy

Complex Coefficient Problems

When dealing with more complex coefficient relationships, your algebraic manipulation skills become crucial. These problems often involve setting up equations where coefficients are related by specific ratios or differences.

Consider when the coefficient of x is 23040 smaller than the coefficient of x² in 2+ax2+ax⁹. You expand the first few terms, identify the relevant coefficients, and set up your equation. The algebra can get messy, but the method stays the same.

Sometimes you'll end up with quadratic equations in your unknown. Use the quadratic formula when factoring isn't obvious, and always check that your answers make sense in the original context.

Don't Panic: These problems look scary but they're just systematic applications of the binomial formula plus some algebra!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students

By signing up you accept Terms of Service and Privacy Policy

Working with Unknown Indices

Unknown indices problems give you information about coefficients and ask you to find the power n. These require you to use the relationship between different terms in the expansion.

If the 3rd and 5th terms have equal coefficients in 1+x1+xⁿ, you set ⁿC₂ = ⁿC₄. Using the factorial formula and simplifying gives you a quadratic equation in n. Remember to check that your answer satisfies any given conditions.

The key insight is that ⁿCᵣ = ⁿCₙ₋ᵣ, which explains why some coefficients can be equal. This symmetry in Pascal's triangle is often the foundation for these problems.

Remember: Pascal's triangle is symmetric - this symmetry is often the key to solving unknown index problems!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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Access to all documents

Improve your grades

Join milions of students

By signing up you accept Terms of Service and Privacy Policy

More Practice with Unknown Indices

Let's solidify your understanding with another approach. When the coefficient of x² in 1+x1+xⁿ equals 55, you use ⁿC₂ = 55. This gives you nn1n-1/2 = 55, leading to n²-n-110 = 0.

For expressions like 1+2x1+2xⁿ where the coefficient of x² is 40, remember that the coefficient involves both the combination and the power of 2. So ⁿC₂ × 2² = 40, giving you ⁿC₂ = 10.

These problems follow the same pattern: identify the relevant term, extract the coefficient using the general formula, set up your equation, and solve. The algebra might vary, but the method is consistent.

Success Strategy: Practice identifying which combination formula you need - that's usually the trickiest part!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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Coefficient Problems with Modified Expressions

When working with expressions like 1+2x1+2xⁿ, don't forget that the coefficient includes powers of the constant. For the x² term, you need ⁿC₂ × (2x)² = ⁿC₂ × 4x².

If this coefficient equals 40, then ⁿC₂ × 4 = 40, so ⁿC₂ = 10. This gives you nn1n-1/2 = 10, leading to n²-n-20 = 0. Factoring gives you n5n-5n+4n+4 = 0.

Since n must be positive, n = 5 is your answer. Always check that your solution makes sense in the context - negative powers or impossible values should make you reconsider your working.

Final Check: Substitute your answer back into the original expansion to verify your coefficient calculation!

Binomial theorem.
(1+x)² = 
1
1
xx
x
x
3
= 1+2x+x²
(1+x)³ = (1+2x+x²) x (1+x)
= x 1 2x x²
1 1 2x x²
xx 2x² x³
= 1+3x+3x²+x³
(1+x)⁴ = (1+3x+3

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This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

Samantha Klich

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Anna

iOS user

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Thomas R

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Just amazing. Let's me revise 10x better, this app is a quick 10/10. I highly recommend it to anyone. I can watch and search for notes. I can save them in the subject folder. I can revise it any time when I come back. If you haven't tried this app, you're really missing out.

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This app has made me feel so much more confident in my exam prep, not only through boosting my own self confidence through the features that allow you to connect with others and feel less alone, but also through the way the app itself is centred around making you feel better. It is easy to navigate, fun to use, and helpful to anyone struggling in absolutely any way.

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