Practise › Questions › The binomial expansion

The binomial expansion questions

Expanding a bracket raised to a high power by repeated multiplication is slow, and the binomial theorem does it in one line using the coefficients from Pascal's triangle. For small values of x the first two or three terms carry almost all the value, which makes the theorem a way of estimating awkward powers by hand.

7 original questions · 22 marks · the binomial expansion notes · Sequences and series

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · warm-up

  Evaluate ⁸C₃ without a calculator, showing the cancellation.

  Worked answer

  8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. M1 for the cancelled quotient, A1 for 56. The 5! wipes out all but three factors of the top, so only 8 × 7 × 6 survives. Computing 8! = 40 320 in full is the long way round and invites arithmetic slips.

• 1 mark · write · warm-up

  Write down the row of Pascal's triangle used for expanding a bracket to the power 5.

  Worked answer

  1, 5, 10, 10, 5, 1, which is six entries for six terms. B1 for the row. Each entry is the sum of the two above it, and the row reads the same in both directions. A power of 5 needs the row starting 1, 5, so counting the apex as row 1 and reaching 1, 4, 6, 4, 1 is the off-by-one to avoid.

• 5 marks

  Expand (3 + 2x)⁴ completely, and check your answer with a substitution.

  Worked answer

  Row 4 gives 1, 4, 6, 4, 1. Build each term with the powers of 3 falling and the powers of 2x climbing: 81, then 4 × 27 × 2x = 216x, then 6 × 9 × 4x² = 216x², then 4 × 3 × 8x³ = 96x³, then 16x⁴. So (3 + 2x)⁴ = 81 + 216x + 216x² + 96x³ + 16x⁴. Setting x = 1 turns the bracket into 5⁴ = 625, and 81 + 216 + 216 + 96 + 16 = 625, so the check passes for the price of one line. B1 for the row of coefficients, M1 for a term with both powers correct, A1 A1 for the middle terms, A1 for the full expansion. Every term needs the 2 raised as well as the x; leaving 2x as 2 in the later terms is the usual collapse.

• 3 marks · find

  Find the first three terms, in ascending powers of x, of (1 + 3x)⁸.

  Worked answer

  1 + ⁸C₁(3x) + ⁸C₂(3x)² = 1 + 24x + 28 × 9x² = 1 + 24x + 252x². M1 for the binomial coefficients with 3x substituted whole, A1 for 24x, A1 for 252x². The replacement 3x goes into the formula whole and gets squared whole. Forgetting to square the 3 gives 84x² and costs the final mark, which is the single commonest loss in this topic.

• 5 marks · find

  Find the first three terms, in ascending powers of x, of (2 + x)⁵, and use them to estimate 2.002⁵.

  Worked answer

  The terms are 2⁵ = 32, then 5 × 2⁴x = 80x, then 10 × 2³x² = 80x². Setting x = 0.002: 32 + 0.16 + 0.00032 = 32.16032. The true value is 32.1603203…, so the truncation is already good to six decimal places; the discarded x³ term is only 40 × 0.002³, three parts in ten million. M1 for a term with the falling power of 2, A1 for 80x, A1 for 80x², M1 for substituting x = 0.002, A1 for 32.16032.

• 4 marks · stretch

  In the expansion of (2 + kx)⁶, where k is a constant, the coefficient of x² is 60. Find the possible values of k.

  Worked answer

  The x² term is ⁶C₂ × 2⁴ × (kx)² = 15 × 16 × k²x² = 240k²x². Setting 240k² = 60 gives k² = ¼, so k = ½ or k = −½. Both are wanted, since only k² appears and the sign cannot be recovered from an even power. At k = ½ the expansion opens 64 + 96x + 60x², which confirms the coefficient. M1 for the x² term of the expansion, A1 for 240k², M1 for solving 240k² = 60, A1 for both values of k. The mark at risk is the 2⁴: the power on the first term drops as the power on the second climbs, and answers that leave it at 2⁶ or omit it entirely reach a different k.

• 2 marks · explain · stretch

  Explain why the coefficients in any row of Pascal's triangle read the same forwards and backwards.

  Worked answer

  The coefficient ⁿC_r counts the ways of choosing which r brackets supply the second letter; choosing those r is the same act as choosing the other n − r to supply the first. So ⁿC_r = ⁿC_n−r, and the row is a palindrome. B1 for choosing r brackets being the same act as choosing the other n − r, B1 for the resulting symmetry of the row.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

Practise the binomial expansion one question at a time

The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.