Binomial expansion (rational powers) Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Binomial expansion (rational powers) questions. See exactly how to solve problems on binomial expansion, rational index, partial fractions.

binomial expansionrational indexpartial fractions
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Expand (1+x)−1\left(1+x\right)^{-1} in ascending powers of xx up to and including the term in x3x^3.

Worked solution

  1. Recall the binomial series for a rational index

    (1+u)−1=1+−1u+−1(−1−1)2!u2+−1(−1−1)(−1−2)3!u3+⋯(1+u)^{-1}=1+-1 u+\frac{-1(-1-1)}{2!}u^{2}+\frac{-1(-1-1)(-1-2)}{3!}u^{3}+\cdots

    For a rational index the expansion is an infinite series, valid when |u|<1.

  2. Identify the index and the small term

    n=−1,u=xn=-1,\quad u=x

    Match the expression to the standard form (1+u)^{n}.

  3. State the expansion up to x^{3}

    1−x+x2−x31 - x + x^{2} - x^{3}

    These are the first four terms of the expansion.

Answer
1−x+x2−x31 - x + x^{2} - x^{3}
Question 2
2 markseasy
Expand (1+x)−2\left(1+x\right)^{-2} in ascending powers of xx up to and including the term in x3x^3.

Worked solution

  1. Recall the binomial series for a rational index

    (1+u)−2=1+−2u+−2(−2−1)2!u2+−2(−2−1)(−2−2)3!u3+⋯(1+u)^{-2}=1+-2 u+\frac{-2(-2-1)}{2!}u^{2}+\frac{-2(-2-1)(-2-2)}{3!}u^{3}+\cdots

    For a rational index the expansion is an infinite series, valid when |u|<1.

  2. Identify the index and the small term

    n=−2,u=xn=-2,\quad u=x

    Match the expression to the standard form (1+u)^{n}.

  3. State the expansion up to x^{3}

    1−2x+3x2−4x31 - 2x + 3x^{2} - 4x^{3}

    These are the first four terms of the expansion.

Answer
1−2x+3x2−4x31 - 2x + 3x^{2} - 4x^{3}
Question 3
2 markseasy
Expand (1+x)12\left(1+x\right)^{\frac{1}{2}} in ascending powers of xx up to and including the term in x3x^3.

Worked solution

  1. Recall the binomial series for a rational index

    (1+u)12=1+12u+12(12−1)2!u2+12(12−1)(12−2)3!u3+⋯(1+u)^{\frac{1}{2}}=1+\frac{1}{2} u+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}u^{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}u^{3}+\cdots

    For a rational index the expansion is an infinite series, valid when |u|<1.

  2. Identify the index and the small term

    n=12,u=xn=\frac{1}{2},\quad u=x

    Match the expression to the standard form (1+u)^{n}.

  3. State the expansion up to x^{3}

    1+12x−18x2+116x31 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{1}{16}x^{3}

    These are the first four terms of the expansion.

Answer
1+12x−18x2+116x31 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{1}{16}x^{3}
Question 4
2 markseasy
Expand (1+x)−12\left(1+x\right)^{- \frac{1}{2}} in ascending powers of xx up to and including the term in x3x^3.

Worked solution

  1. Recall the binomial series for a rational index

    (1+u)−12=1+−12u+−12(−12−1)2!u2+−12(−12−1)(−12−2)3!u3+⋯(1+u)^{- \frac{1}{2}}=1+- \frac{1}{2} u+\frac{- \frac{1}{2}(- \frac{1}{2}-1)}{2!}u^{2}+\frac{- \frac{1}{2}(- \frac{1}{2}-1)(- \frac{1}{2}-2)}{3!}u^{3}+\cdots

    For a rational index the expansion is an infinite series, valid when |u|<1.

  2. Identify the index and the small term

    n=−12,u=xn=- \frac{1}{2},\quad u=x

    Match the expression to the standard form (1+u)^{n}.

  3. State the expansion up to x^{3}

    1−12x+38x2−516x31 - \frac{1}{2}x + \frac{3}{8}x^{2} - \frac{5}{16}x^{3}

    These are the first four terms of the expansion.

Answer
1−12x+38x2−516x31 - \frac{1}{2}x + \frac{3}{8}x^{2} - \frac{5}{16}x^{3}
Question 5
2 markseasy
Find the coefficient of x2x^{2} in the expansion of (1+x)−1\left(1+x\right)^{-1}.

Worked solution

  1. Recall the binomial series for a rational index

    (1+u)−1=1+−1u+−1(−1−1)2!u2+−1(−1−1)(−1−2)3!u3+⋯(1+u)^{-1}=1+-1 u+\frac{-1(-1-1)}{2!}u^{2}+\frac{-1(-1-1)(-1-2)}{3!}u^{3}+\cdots

    For a rational index the expansion is an infinite series, valid when |u|<1.

  2. Identify the index and the small term

    n=−1,u=xn=-1,\quad u=x

    Match the expression to the standard form (1+u)^{n}.

  3. State the coefficient of x^{2}

    coefficient of x2=1\text{coefficient of }x^{2}=1

    Read off the coefficient of x^{2} from the expansion.

Answer
11

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