Questions on the general binomial expansion

Question 1

The binomial expansion of (1 + x)ⁿ stops after n + 1 terms when n is a positive whole number, but never stops when n = −1. Explain the difference, and state the validity condition for the never-ending case.

Accepted Answer

For whole n the coefficient n(n − 1)…(n − r + 1)/r! eventually picks up the factor zero, and every term after that dies. For n = −1 the factors run −1, −2, −3, … and never hit zero, so the series continues for ever. It then only converges for |x| < 1, and that condition has to be quoted with the expansion. B1 B1 for the two explanations, B1 for |x| < 1.

Question 2

Expand (1 + x)⁻¹ in ascending powers of x up to the x³ term, and state the validity.

Accepted Answer

With n = −1 the expansion is 1 − x + x² − x³ + …, valid for |x| < 1. M1 for the expansion, A1 for 1 − x + x² − x³, B1 for the validity. The alternating signs come from the coefficient factors −1, −2, −3 dividing by 1!, 2! and 3!, each ratio working out at ±1. This is the geometric series for 1/(1 + x) in binomial clothing, first term 1 and common ratio −x, and the same |r| < 1 condition appears from both directions.

Question 3

State the range of values of x for which the binomial expansion of (1 + x/5)⁻³ is valid, and the range for which the expansion of (1 − 5x)^2/3 is valid. State which of the two expansions is valid at x = 0.3.

Accepted Answer

The series converges when the quantity that replaces x in the formula has modulus below 1. For the first expansion that quantity is x/5, so |x/5| < 1 and |x| < 5. B1 for |x| < 5. For the second it is −5x, so |−5x| < 1 and |x| < 1/5. B1 for |x| < 1/5. At x = 0.3 the first condition holds and the second fails, since 0.3 is larger than 0.2, so only the expansion of (1 + x/5)⁻³ is valid there. B1 for the comparison. The window is set by the coefficient on x alone, and the two here are reciprocals: dividing x by 5 widens the window to 5, while multiplying x by 5 narrows it to 1/5. The sign of that coefficient and the value of the exponent do not enter it.

Question 4

Find the coefficient of x³ in the binomial expansion of (1 + x)^1/3.

Accepted Answer

The coefficient of x³ in (1 + x)ⁿ is n(n − 1)(n − 2)/3!. M1 for that coefficient with n = 1/3. The three factors are 1/3, −2/3 and −5/3, so the coefficient is (1/3)(−2/3)(−5/3)/6 = (10/27)/6. A1 for (10/27)/6. That comes to 5/81. A1 for 5/81. Every factor after the first is negative when n lies between 0 and 1, so two of them here multiply to a positive: the signs have to be carried one factor at a time rather than assumed to alternate.

Question 5

Expand (1 + 4x)^1/2 in ascending powers of x up to the x² term, and state the range of x for which the expansion is valid.

Accepted Answer

Apply the formula with n = ½ and 4x in place of x: 1 + ½(4x) + [½ × (−½)/2](4x)² = 1 + 2x − 2x². Validity needs |4x| < 1, so |x| < ¼. M1 for the formula with n = ½, A1 for the x term, A1 for the x² term, B1 for the validity. The replacement goes in whole and gets squared whole, and the validity window shrinks by the same factor of 4. Squaring only the x reaches −x²/8.

Question 6

Expand (1 + 2x)⁻² up to the x² term, state the validity, and use the expansion to estimate 1.02⁻².

Accepted Answer

With n = −2 and 2x in place of x: 1 + (−2)(2x) + [(−2)(−3)/2](2x)² = 1 − 4x + 12x², valid for |x| < ½. Setting x = 0.01: 1 − 0.04 + 0.0012 = 0.9612, against a true value of 0.961169…: agreement to four decimal places from three terms. M1 for the expansion, A1 for 1 − 4x + 12x², B1 for the validity, M1 for setting x = 0.01, A1 for 0.9612.

Question 7

Expand (4 + x)^1/2 up to the x² term, and state the validity.

Accepted Answer

The bracket does not start with 1, so factor the 4 out: (4 + x)^1/2 = 2(1 + x/4)^1/2. Expanding the inner bracket: 1 + x/8 − x²/128 + …, and doubling gives 2 + x/4 − x²/64. Validity needs |x/4| < 1, so |x| < 4. M1 for taking the 4 outside, A1 for the factor 2, M1 for expanding the inner bracket, A1 for 2 + x/4 − x²/64, B1 for the validity. The factored-out 4 comes through as 4^1/2 = 2, not 4. As a sanity check, x = 0.2 gives 2.049375 against a true √4.2 = 2.0493901…, so three terms are already worth four decimal places.

Question 8

Expand (8 + 3x)^1/3 in ascending powers of x up to the x² term, and state the range of x for which the expansion is valid.

Accepted Answer

Factor the 8 out first: (8 + 3x)^1/3 = 8^1/3(1 + 3x/8)^1/3 = 2(1 + 3x/8)^1/3. With n = 1/3 and 3x/8 in place of x, the inner bracket expands to 1 + (1/3)(3x/8) + [(1/3)(−2/3)/2](3x/8)² = 1 + x/8 − x²/64, and doubling gives 2 + x/4 − x²/32. Validity needs |3x/8| < 1, so |x| < 8/3. M1 for taking the 8 out as 8^1/3 = 2, M1 for the expansion with n = 1/3, A1 for 2 + x/4 − x²/32, B1 for the validity. The factored constant leaves as a cube root, not as 8, and the fractional n makes both coefficient factors fractions: (1/3)(−2/3)/2 = −1/9, and (−1/9)(9/64) = −1/64. A check at x = 0.1 gives 2.0246875 against a true 8.3^1/3 = 2.0246939…, four decimal places from three terms.

Question 9

Expand (1 − 2x)^1/2 in ascending powers of x up to the x² term, and state the validity. Given that 0.98 = 49 × 2/100, choose a suitable value of x and use the expansion to estimate √2 to 5 decimal places.

Accepted Answer

With n = ½ and −2x in place of x: 1 + ½(−2x) + [½ × (−½)/2](−2x)² = 1 − x − x²/2, valid for |2x| < 1, so |x| < ½. At x = 0.01 the bracket is 0.98, and since 0.98 = 49 × 2/100, √0.98 = 7√2/10, so √2 = (10/7)√0.98. The truncation gives √0.98 ≈ 1 − 0.01 − 0.00005 = 0.98995, so √2 ≈ 9.8995/7 = 1.41421 to 5 decimal places. M1 for the expansion with −2x kept whole, A1 for 1 − x − x²/2, B1 for the validity, M1 for √0.98 = 7√2/10, M1 for x = 0.01 substituted and rearranged, A1 for 1.41421. The true value is 1.4142136…, so all five places are right: the discarded term −x³/2 is worth only 7 × 10⁻⁷ after the factor of 10/7, and that is the whole of the error.

Question 10

The binomial expansion of (1 + kx)⁻³ in ascending powers of x begins 1 + 9x, where k is a constant. Find the value of k, and hence find the x² and x³ terms of the expansion.

Accepted Answer

With n = −3 and kx in place of x, the x term is (−3)(kx) = −3kx, so −3k = 9 and k = −3. M1 for −3k = 9, A1 for k = −3. The replacement is therefore −3x. The x² term is [(−3)(−4)/2](−3x)² = 6 × 9x² = 54x². A1 for 54x². The x³ term is [(−3)(−4)(−5)/6](−3x)³ = (−10)(−27x³) = 270x³. A1 for 270x³. The replacement −3x is raised whole at every stage, sign included: cubing it keeps the minus, which then meets the negative coefficient −10 and leaves a positive term. The factors −3, −4, −5 never reach zero, which is why this expansion has no last term.

Question 11

Expand (1 + x)^1/2 in ascending powers of x up to the x³ term. By evaluating those four terms at x = 3, show that the truncated series does not estimate √4, and state what has gone wrong.

Accepted Answer

With n = ½ the formula gives 1 + ½x + [½ × (−½)/2]x² + [½ × (−½) × (−3/2)/6]x³ = 1 + x/2 − x²/8 + x³/16. M1 for the formula with n = ½, A1 for the four terms. At x = 3 the bracket is 4, and the four terms come to 1, 1.5, −1.125 and 1.6875. M1 for evaluating them at x = 3. Their total is 3.0625, while √4 = 2, so the truncation is nowhere near the value it is supposed to estimate. A1 for 3.0625 against 2. The condition |x| < 1 fails at x = 3, and the consequence is visible in the terms themselves: the fourth is larger than the third, so they are not shrinking and the series does not converge. B1 for the validity condition failing. Adding further terms would make matters worse rather than better, and a truncation is only worth quoting inside the interval where the full series converges.

Question 12

Given that |x| < ¼, expand (1 + 2x)/(1 − 4x)^1/2 in ascending powers of x up to and including the x² term.

Accepted Answer

Treat the denominator as a negative power: the expression is (1 + 2x)(1 − 4x)^−1/2. Expanding the second factor with n = −½ and −4x in place of x gives 1 + (−½)(−4x) + [(−½)(−3/2)/2](−4x)² = 1 + 2x + 6x². Now multiply by (1 + 2x), keeping only terms as far as x²: the x term is 2x + 2x = 4x, and the x² term is 6x² + 4x² = 10x². So the expansion is 1 + 4x + 10x². M1 for writing the denominator as a negative power, M1 for expanding it, A1 for 1 + 2x + 6x², M1 for multiplying out as far as x², A1 for the x term, A1 for the x² term. Two traps sit here. The minus sign inside the bracket has to survive the squaring, since (−4x)² is +16x²; and the multiplying-out stage must not drop the cross term 2x × 2x. The validity |4x| < 1 is exactly the |x| < ¼ the question hands you, which is a hint that the −4x has been read correctly.

Question 13

Given that (2 − x)/((1 + x)(1 − 2x)) = A/(1 + x) + B/(1 − 2x), find A and B. Hence expand (2 − x)/((1 + x)(1 − 2x)) in ascending powers of x up to the x² term, and state the range of x for which the expansion is valid.

Accepted Answer

Multiply up: 2 − x = A(1 − 2x) + B(1 + x). Putting x = −1 gives 3 = 3A, so A = 1; putting x = ½ gives 3/2 = (3/2)B, so B = 1. The fraction is 1/(1 + x) + 1/(1 − 2x). Expand each piece: (1 + x)⁻¹ = 1 − x + x² − …, valid for |x| < 1, and (1 − 2x)⁻¹ = 1 + 2x + 4x² + …, valid for |2x| < 1, that is |x| < ½. Adding term by term gives 2 + x + 5x². Both windows must hold at once, so the expansion is valid only on the tighter, |x| < ½. M1 for multiplying up, A1 for A = 1 and B = 1, M1 for expanding either bracket binomially, A1 for 1 − x + x², A1 for 1 + 2x + 4x², A1 for 2 + x + 5x², B1 for |x| < ½ as the surviving window. A decimal check at x = 0.1: the original fraction is 1.9/(1.1 × 0.8) = 2.1591 while the truncation gives 2.15, agreeing to two decimal places, with the discarded terms from 7x³ = 0.007 onwards carrying most of the gap.

Question 14

Show that in the binomial expansion of (1 − x)⁻² in ascending powers of x the coefficient of x^r is r + 1. Write down the first five terms, and state the range of x for which the expansion is valid.

Accepted Answer

The general term of (1 + y)ⁿ is [n(n − 1)…(n − r + 1)/r!]y^r, and here n = −2 with y = −x. M1 for the general coefficient with n = −2. Its numerator runs (−2)(−3)…(−(r + 1)), which is r factors, so it equals (−1)^r × 2 × 3 × … × (r + 1) = (−1)^r(r + 1)!. M1 for the numerator as (−1)^r(r + 1)!. Dividing by r! leaves (−1)^r(r + 1), and the term also carries (−x)^r = (−1)^rx^r, so the two sign factors multiply to (−1)^2r = 1. A1 for the signs cancelling. The coefficient of x^r is therefore r + 1. A1 for r + 1. Taking r = 0, 1, 2, 3 and 4, the expansion opens 1 + 2x + 3x² + 4x³ + 5x⁴. B1 for the five terms. Validity needs |−x| < 1, that is |x| < 1. B1 for the validity. The coefficients grow without limit, so it is the powers of x that have to do the shrinking, and they only manage it inside that interval.

Question 15

The binomial expansion of (1 + ax)ⁿ in ascending powers of x begins 1 − 12x + 90x², where a and n are constants. Find the values of a and n, and state the range of x for which the expansion is valid.

Accepted Answer

Matching the x term, na = −12. M1 for na = −12. Matching the x² term, [n(n − 1)/2]a² = 90. M1 for the second equation. From the first, a = −12/n, so a² = 144/n² and the second becomes n(n − 1)(144/n²)/2 = 72(n − 1)/n = 90. M1 for eliminating a. Then 72n − 72 = 90n, so 18n = −72 and n = −4. A1 for n = −4. Back-substituting, a = −12/(−4) = 3. A1 for a = 3. Validity needs |3x| < 1, that is |x| < 1/3. B1 for the validity. Checking, (1 + 3x)⁻⁴ opens 1 + (−4)(3x) + [(−4)(−5)/2](3x)² = 1 − 12x + 90x², as it should. Guessing n from the size of the printed coefficients would not work, since a positive whole n gives coefficients that grow for the first few terms as well.