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The general binomial expansion questions
Let the exponent be negative or fractional and the binomial theorem keeps going, but the expansion never stops. What you get is an infinite series that is valid only inside a window of x-values. Learn where the window is and the series will compute square roots and reciprocals to whatever accuracy you want.
6 original questions · 26 marks · the the general 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.
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.
Worked 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 quoting that condition is a mark in its own right on every question of this type. B1 B1 for the two explanations, B1 for |x| < 1.Expand (1 + x)−1 in ascending powers of x up to the x3 term, and state the validity.
Worked 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.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.
Worked 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. Answers that square only the x reach −x2/8 and lose two marks together.Expand (1 + 2x)−2 up to the x2 term, state the validity, and use the expansion to estimate 1.02−2.
Worked 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.Expand (4 + x)1/2 up to the x2 term, and state the validity.
Worked 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. Forgetting that the factored-out 4 comes through as 41/2 = 2, not 4, is where this question usually goes wrong. 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.Given that |x| < ¼, expand (1 + 2x)/(1 − 4x)1/2 in ascending powers of x up to and including the x2 term.
Worked 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.
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