Practice Maths

The Binomial Theorem

Key Terms

A binomial is an expression with two terms, e.g. (x + y) or (2x − 3).
The Binomial Theorem gives a formula for expanding (x + y)n for any positive integer n.
The general term (the (r+1)th term) is: Tr+1 = nCr xn−r yr.
Coefficients come from Pascal’s triangle or the combination formula nCr = n!/(r!(n−r)!).
The powers of the first term decrease from n to 0; powers of the second term increase from 0 to n.
There are always n + 1 terms in the expansion of (x + y)n.
Simplify each term carefully when y is negative (e.g. (a − b)n) — the signs alternate.
★ Formula Sheet — Binomial Theorem (this formula is given on your QCAA formula sheet):
(x + y)n = nC0 xn + nC1 xn−1y + nC2 xn−2y2 + … + nCr xn−ryr + … + yn
General term: Tr+1 = nCr xn−r yr
nExpansion of (x + y)n
1x + y
2x² + 2xy + y²
3x³ + 3x²y + 3xy² + y³
4x&sup4; + 4x³y + 6x²y² + 4xy³ + y&sup4;
Hot Tip For (a − b)n, substitute y = −b into the theorem. Odd-powered terms of b become negative, even-powered terms are positive — the signs alternate starting with + for the first term.

Worked Example 1 — Full Expansion

Question: Expand (x + 2)4.

Using the theorem with y = 2, n = 4:

= 4C0x4 + 4C1x3(2) + 4C2x2(2)2 + 4C3x(2)3 + 4C4(2)4

= 1·x4 + 4·2x3 + 6·4x2 + 4·8x + 1·16

Answer: x4 + 8x3 + 24x2 + 32x + 16

Worked Example 2 — Finding a Specific Term

Question: Find the term containing x3 in the expansion of (2x − 3)5.

Step 1: Use Tr+1 = 5Cr(2x)5−r(−3)r. We need the power of x to be 3, so 5 − r = 3, giving r = 2.

Step 2: T3 = 5C2(2x)3(−3)2 = 10 × 8x3 × 9 = 720x3.

Answer: 720x3

Expanding Without Multiplying Repeatedly

The binomial theorem gives a direct formula for the coefficients in the expansion of (x + y)n without needing to multiply out n copies of (x + y) step by step. This matters enormously for large n: expanding (x + y)8 by repeated multiplication would require seven multiplications and considerable algebra. The binomial theorem gives all nine terms directly.

The combinatorial explanation: in the product (x + y)(x + y)…(x + y) [n factors], each term is formed by choosing x or y from each factor. To get the term xn-ryr, you must choose y exactly r times from the n factors. The number of ways to make this choice is nCr, which becomes the coefficient. This is why binomial coefficients appear naturally — they are not an arbitrary rule but a direct consequence of how products work.

The General Term Formula: Tr+1 = nCr xn-r yr

The subscript notation is critical and often confusing. Tr+1 means the (r+1)th term of the expansion. When r = 0, we get T1 (the first term). When r = 1, we get T2 (the second term). The reason we use r+1 rather than r as the term number is that r is the power of y, which starts at 0 (not 1) in the first term.

In the expansion of (a + b)n with more complex expressions, substitute a for x and b for y. For example, in (2x + 3)5: Tr+1 = 5Cr(2x)5-r(3)r. The powers of (2x) and (3) must both be included — not just the power of x.

Finding Specific Terms: Setting Up and Solving for r

To find the term containing xk in an expansion, write the general term, identify what power of x it contains as a function of r, set that equal to k, and solve for r. The value of r must be a non-negative integer between 0 and n; if no such integer exists, there is no such term.

For the term containing x3 in (2x − 3)5: Tr+1 = 5Cr(2x)5-r(−3)r. The power of x is 5 − r. Set 5 − r = 3, giving r = 2. So T3 = 5C2(2x)3(−3)2 = 10 × 8x³ × 9 = 720x³.

Special Cases: Sums and Differences

For (x + 1)n, simply set y = 1. This gives integer coefficients equal to the nth row of Pascal’s triangle: nC0xn + nC1xn-1 + … + nCn.

For (x − 1)n, set y = −1. The signs alternate: terms with odd powers of y (which is −1) become negative. The pattern is + − + − + … In general, for (a − b)n, the signs alternate starting from + for the first term.

Adding (x + 1)n and (x − 1)n cancels all odd-powered terms; subtracting cancels all even-powered terms. These combinations appear in exam problems asking for specific sums.

Approximations Using the Binomial Theorem

For small values of x, the binomial theorem gives excellent approximations: (1 + x)n ≈ 1 + nx for |x| << 1. This is the first-order approximation, keeping only the first two terms. The error comes from the neglected terms nC2x² + …, which are tiny when x is small.

For example, (1.02)10 = (1 + 0.02)10 ≈ 1 + 10(0.02) = 1.2. The exact value is 1.2190…, so this gives a good first approximation. A better approximation includes the x² term: 1 + 10(0.02) + 45(0.0004) = 1.218, much closer to the exact value.

Exam Tip: The general term is Tr+1, not Tr. This means: the first term (r = 0) is T1, the second term (r = 1) is T2, and so on. A very common error is to compute Tr = nCr… instead of Tr+1 = nCr…, which shifts every term by one position. Always double-check which term you are computing by testing r = 0 and verifying you get the first term of the expansion.
Exam Tip: When finding the constant term (the term independent of x), set the total power of x in Tr+1 equal to zero and solve for r. In (x + 1/x)6: Tr+1 = 6Crx6-r(x-1)r = 6Crx6-2r. Set 6 − 2r = 0, so r = 3. The constant term is 6C3 = 20.

Mastery Practice

See Answers ➔

  1. Write out each expansion using the binomial theorem. Fluency

    1. (a) (x + 1)3
    2. (b) (x + y)4
    3. (c) (a − b)3
    4. (d) (1 + x)5
    5. (e) (x − 2)4

  2. Expand and simplify. Fluency

    1. (a) (2x + 1)3
    2. (b) (x + 3)4
    3. (c) (2 − x)4
    4. (d) (3x + 2)3

  3. Find the general term Tr+1 for each expansion, and hence find the specified term. Fluency

    1. (a) (x + 2)6: find T4.
    2. (b) (1 + x)7: find T5.
    3. (c) (x − 1)8: find T3.
    4. (d) (2x + 3)5: find T2.

  4. Find the term containing the given power of x. Fluency

    1. (a) Coefficient of x3 in (1 + x)7.
    2. (b) Coefficient of x2 in (x + 3)5.
    3. (c) Coefficient of x4 in (2x − 1)6.
    4. (d) Constant term in (x + 2/x)6. (Hint: find r such that the power of x is 0.)

  5. Expand using the binomial theorem and collect like terms. Fluency

    1. (a) (1 + x)4 + (1 − x)4
    2. (b) (1 + x)4 − (1 − x)4

  6. Applying the theorem. Understanding

    Evaluating numerical expressions.
    1. (a) Use the binomial theorem to expand (1 + 0.1)4 and hence find an approximate value for 1.14 correct to 4 decimal places.
    2. (b) Use (1 − 0.01)5 to find an approximate value for 0.995 correct to 5 decimal places.
    3. (c) Find the exact value of (1 + √2)4 + (1 − √2)4.

  7. Greatest coefficient and largest term. Understanding

    Investigating the expansion of (1 + x)n.
    1. (a) Write out all terms of (1 + x)6. State the term with the greatest coefficient.
    2. (b) For the expansion of (2 + x)5, find the value of x (if any) for which T3 = T4.

  8. Connection to identities. Understanding

    Known expansions.
    1. (a) Use the binomial theorem with x = 1 and y = 1 to show that nC0 + nC1 + … + nCn = 2n.
    2. (b) Use x = 1 and y = −1 to show that the alternating sum nC0 − nC1 + … = 0 for n ≥ 1.

  9. Finding unknown coefficients. Problem Solving

    Working backwards from a known term.
    1. (a) The coefficient of x3 in (1 + kx)6 is 160. Find k.
    2. (b) The coefficient of x2 in the expansion of (2 + ax)4 is 96. Find a.
    3. (c) In the expansion of (1 + px)n, the coefficient of x is 12 and the coefficient of x2 is 60. Find p and n.

  10. Multinomial-style challenge. Problem Solving

    Combining two expansions.
    1. (a) Find the coefficient of x5 in the expansion of (1 + x)4(1 + x)3. (Hint: combine first, then expand, OR expand each separately and multiply.)
    2. (b) Find the coefficient of x3 in the expansion of (1 + x + x2)(1 + x)4.
    3. (c) Show that the middle term of (1 + x)2n has coefficient 2nCn.