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How does the binomial theorem let us expand a power of a bracket like (a plus b) to the n without multiplying it out by hand?

Use the binomial theorem to expand (a plus b) to the power n for a positive integer n, using binomial coefficients

A focused answer to the N(A)-Level Additional Mathematics outcome on the binomial theorem. Expand a power of a bracket using binomial coefficients, with the help of Pascal's triangle and the nCr notation.

Generated by Claude Opus 4.88 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to expand a power of a bracket, (a+b)n(a + b)^n for a positive whole number nn, using the binomial theorem rather than multiplying the bracket out many times. You need the binomial coefficients (from Pascal's triangle or the (nr)\binom{n}{r} notation) and the pattern of decreasing and increasing powers. This is the foundation for finding a single specific term, covered in the next dot point.

The answer

The pattern of an expansion

When you expand (a+b)n(a + b)^n, every term is a coefficient times a power of aa times a power of bb. As you move along the expansion:

  • the power of aa decreases from nn down to 00,
  • the power of bb increases from 00 up to nn,
  • the two powers in each term always add up to nn.

There are n+1n + 1 terms in total.

Binomial coefficients from Pascal's triangle

The coefficients are the numbers in Pascal's triangle, where each number is the sum of the two above it:

  • Row n=2n = 2: 1, 2, 11,\ 2,\ 1
  • Row n=3n = 3: 1, 3, 3, 11,\ 3,\ 3,\ 1
  • Row n=4n = 4: 1, 4, 6, 4, 11,\ 4,\ 6,\ 4,\ 1
  • Row n=5n = 5: 1, 5, 10, 10, 5, 11,\ 5,\ 10,\ 10,\ 5,\ 1

For small nn, reading off the right row is the fastest method.

The general formula

For larger nn, the coefficients are written (nr)\binom{n}{r} (read "nn choose rr"), and the theorem is:

(a+b)n=(n0)an+(n1)an1b+(n2)an2b2++(nn)bn(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n

where (nr)\binom{n}{r} can be found on your calculator (often labelled nCrnCr).

Expanding a numerical bracket

When aa or bb is a number, raise it to the right power in each term and simplify. Keep track of powers carefully, for example (2+x)3(2 + x)^3 uses 23,22,21,202^3, 2^2, 2^1, 2^0 as the power of aa falls.

Examples in context

Example 1. Approximating a value. Expanding (1+x)n(1 + x)^n for small xx and keeping only the first few terms gives a quick approximation, for example (1.02)51+5(0.02)=1.10(1.02)^5 \approx 1 + 5(0.02) = 1.10. The early terms of a binomial expansion are the basis of such estimates.

Example 2. Probability with repeated trials. The coefficients (nr)\binom{n}{r} count the ways an outcome can happen across nn trials, which is why the same binomial numbers appear in the binomial probability distribution. Expanding the bracket and counting arrangements are two sides of the same idea.

Try this

Q1. Write down the coefficients for the expansion of a bracket to the power 55. [1 mark]

  • Cue. From Pascal's triangle: 1,5,10,10,5,11, 5, 10, 10, 5, 1.

Q2. Expand (1+x)3(1 + x)^3. [2 marks]

  • Cue. 1+3x+3x2+x31 + 3x + 3x^2 + x^3.

Q3. Expand (12x)2(1 - 2x)^2. [2 marks]

  • Cue. 14x+4x21 - 4x + 4x^2, since (2x)2=4x2(-2x)^2 = 4x^2.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksExpand (1+x)4(1 + x)^4 in ascending powers of xx.
Show worked answer →

The coefficients from row 4 of Pascal's triangle are 1,4,6,4,11, 4, 6, 4, 1.

So (1+x)4=1+4x+6x2+4x3+x4(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4.

What markers reward: using the correct row of coefficients, attaching increasing powers of xx, and giving every term from x0x^0 to x4x^4.

Original4 marksExpand (2+x)3(2 + x)^3 completely.
Show worked answer →

Coefficients for power 3 are 1,3,3,11, 3, 3, 1. Apply them to decreasing powers of 22 and increasing powers of xx:

(2+x)3=1(2)3+3(2)2x+3(2)x2+1x3=8+12x+6x2+x3(2 + x)^3 = 1(2)^3 + 3(2)^2 x + 3(2)x^2 + 1 x^3 = 8 + 12x + 6x^2 + x^3.

What markers reward: pairing the coefficients with the correct powers of 22 and xx, evaluating 23=82^3 = 8 and 3×22=123 \times 2^2 = 12, and giving the full expansion.

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