Questions on the binomial expansion

Question 1

Evaluate ⁸C₃ without a calculator, showing the cancellation.

Accepted 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.

Question 2

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

Accepted 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.

Question 3

Write down, in ascending powers of x, the first three terms of the expansion of (1 + x)⁷, and state the coefficient of x⁵.

Accepted Answer

Row 7 of Pascal's triangle is 1, 7, 21, 35, 35, 21, 7, 1, so the expansion opens 1 + 7x + 21x². B1 for the first three terms. The coefficient of x⁵ is ⁷C₅ = 21. B1 for 21. It matches the coefficient of x², because ⁷C₅ and ⁷C₂ count the same choices from opposite ends of the row, so a coefficient late in a row can be read off early in it.

Question 4

Find the coefficient of x⁴ in the expansion of (2 − 3x)⁶.

Accepted Answer

The x⁴ term is ⁶C₄ × 2² × (−3x)⁴. M1 for the term with both powers correct and the whole of −3x raised. That is 15 × 4 × 81x⁴. A1 for 15 × 4 × 81. So the coefficient is 4860. A1 for 4860. The power on the 2 is 6 − 4 = 2 rather than 6, since the two powers in a term add to 6; and the minus sign is raised with the 3, disappearing here because the power is even.

Question 5

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

Accepted 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. 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.

Question 6

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

Accepted 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 whole term 3x is squared, so the coefficient carries 3² = 9; squaring only x gives 84x².

Question 7

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

Accepted 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, about three ten-millionths. 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.

Question 8

In the expansion of (1 + x)ⁿ, where n is a positive integer, the coefficient of x² is 66. Find the value of n, and write down the coefficient of x³.

Accepted Answer

The coefficient of x² is ⁿC₂ = n(n − 1)/2, so n(n − 1)/2 = 66 and n(n − 1) = 132. M1 for n(n − 1)/2 = 66. Rearranged, n² − n − 132 = 0, which factorises as (n − 12)(n + 11) = 0. M1 for the quadratic. A power has to be a positive integer, so the root n = −11 is discarded and n = 12. A1 for n = 12. The coefficient of x³ is then ¹²C₃ = (12 × 11 × 10)/6 = 220. B1 for 220. Two consecutive whole numbers with product 132 can also be spotted as 12 and 11, which saves forming the quadratic when the numbers are this small.

Question 9

Find the coefficient of x² in the expansion of (1 + 2x)(1 − 3x)⁴.

Accepted Answer

Expand the bracket carrying the power first, as far as x²: 1 + ⁴C₁(−3x) + ⁴C₂(−3x)² = 1 − 12x + 54x². M1 for the first three terms with −3x kept whole, A1 for 1 − 12x + 54x². An x² term of the product can be built two ways: 1 × 54x² and 2x × (−12x). M1 for collecting both contributions. So the coefficient is 54 − 24 = 30. A1 for 30. Terms beyond x² in the expansion cannot reach x² in the product, so three terms are enough.

Question 10

Expand (1 + x)⁵ and (1 − x)⁵ in ascending powers of x, and hence show that (1 + x)⁵ + (1 − x)⁵ = 2 + 20x² + 10x⁴. Use the result to find the exact value of 1.1⁵ + 0.9⁵.

Accepted Answer

Row 5 gives 1, 5, 10, 10, 5, 1, so (1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵. B1 for that expansion. Replacing x by −x changes the sign of every odd power: (1 − x)⁵ = 1 − 5x + 10x² − 10x³ + 5x⁴ − x⁵. M1 for the alternating signs. Adding the two lines, the odd powers cancel in pairs and the even ones double. M1 for adding, A1 for 2 + 20x² + 10x⁴. Setting x = 0.1 makes the two brackets 1.1 and 0.9, so 1.1⁵ + 0.9⁵ = 2 + 20 × 0.01 + 10 × 0.0001 = 2.201. A1 for 2.201. Nothing is discarded on the way, so the value is exact rather than an estimate: the identity holds for every x.

Question 11

The coefficient of x² in the expansion of (1 + ax)⁴ is equal to the coefficient of x³ in the expansion of (1 + ax)⁶. Given that a ≠ 0, find the value of a.

Accepted Answer

The x² term of the first expansion is ⁴C₂(ax)² = 6a²x². M1 for the x² term with ax squared whole, A1 for 6a². The x³ term of the second is ⁶C₃(ax)³ = 20a³x³, so the condition reads 6a² = 20a³. M1 for equating the two coefficients. Dividing by a², which is allowed because a ≠ 0, leaves 6 = 20a and a = 3/10. A1 for a = 3/10.

Question 12

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

Accepted 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 power to watch 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.

Question 13

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

Accepted 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.

Question 14

In the expansion of (1 + x)ⁿ, where n is a positive integer, use the substitutions x = 1 and x = −1 to show that the coefficients of row n add to 2ⁿ, and that the coefficients of the even powers of x have the same total as the coefficients of the odd powers. State that common total when n = 8.

Accepted Answer

Putting x = 1 turns the bracket into 2ⁿ and every power of x into 1, so the expansion becomes the plain sum of its coefficients: ⁿC₀ + ⁿC₁ + … + ⁿC_n = 2ⁿ. M1 for the substitution x = 1, A1 for the total 2ⁿ. Putting x = −1 makes the bracket 0, while (−1)^r attaches a minus sign to every odd power, so ⁿC₀ − ⁿC₁ + ⁿC₂ − … = 0. M1 for the substitution x = −1. A total of zero says the coefficients carrying a minus sign cancel those carrying a plus, so the odd-power coefficients and the even-power coefficients have equal totals. A1 for the two totals being equal. Between them the two results make each total half of 2ⁿ, that is 2ⁿ⁻¹, so at n = 8 each comes to 128. B1 for 128. The second substitution needs n to be at least 1 for the bracket to vanish: at n = 0 the row is a single 1, its alternating total is 1, and there are no odd powers for it to balance.

Question 15

Prove that ⁿC_r−1 + ⁿC_r = ⁿ⁺¹C_r for integers n and r with 1 ≤ r ≤ n, and state what the identity proves about Pascal's triangle.

Accepted Answer

Write both coefficients as factorials: ⁿC_r−1 = n!/[(r − 1)!(n − r + 1)!] and ⁿC_r = n!/[r!(n − r)!]. M1 for both in factorial form. The common denominator is r!(n − r + 1)!: the first fraction needs its (r − 1)! turned into r!, so multiply its top and bottom by r, and the second needs its (n − r)! turned into (n − r + 1)!, so multiply its top and bottom by n − r + 1. M1 for reaching the common denominator. The numerators are then n! × r and n! × (n − r + 1). A1 for both numerators. Adding them, n!(r + n − r + 1) = n!(n + 1) = (n + 1)!. M1 for the numerator collapsing to (n + 1)!. So the sum is (n + 1)!/[r!(n + 1 − r)!], which is ⁿ⁺¹C_r. A1 for the result. Row n + 1 of Pascal's triangle holds the coefficients ⁿ⁺¹C_r, so the identity is the proof that every entry of the triangle is the sum of the two entries above it. B1 for that statement. The range 1 ≤ r ≤ n keeps every factorial defined; the two entries at the ends of a row have only one entry above them and are 1 by the definition of ⁿC₀ and ⁿC_n.