A-Level Binomial expansion (rational powers) Practice Questions

Free A-Level Binomial expansion (rational powers) practice questions with full step-by-step worked solutions. Covers binomial expansion, rational index, partial fractions. Practise exam-style problems and check your method.

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.
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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
In the expansion of (1+x)12\left(1+x\right)^{\frac{1}{2}}, what is the coefficient of x1x^{1}?
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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. Select the correct coefficient

    coefficient of x1=12\text{coefficient of }x^{1}=\frac{1}{2}

    The coefficient of x^{1} is read from the expansion.

Answer
12\frac{1}{2}
Question 3
3 marksintermediate
Which of the following gives the first three terms of the expansion of (1+x)−2\left(1+x\right)^{-2} in ascending powers of xx?
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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. Substitute the small term into the series

    1+−2(x)+−2(−2−1)2!(x)2+−2(−2−1)(−2−2)3!(x)31+-2\left(x\right)+\frac{-2(-2-1)}{2!}\left(x\right)^{2}+\frac{-2(-2-1)(-2-2)}{3!}\left(x\right)^{3}

    Replace u by the actual term ready to simplify.

  4. State the constant term

    11

    The first term of (1+u)^{n} is always 1.

  5. Compute the coefficient of u

    −2-2

    The linear coefficient equals the index n.

  6. Select the correct first three terms

    1−2x+3x21 - 2x + 3x^{2}

    The first three terms come from the coefficients of 1, x and x^{2}.

Answer
1−2x+3x21 - 2x + 3x^{2}
Question 4
5 markshard
Which of the following gives the first three terms of the expansion of (1+x)−3\left(1+x\right)^{-3} in ascending powers of xx?
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Worked solution

  1. Recall the binomial series for a rational index

    (1+u)−3=1+−3u+−3(−3−1)2!u2+−3(−3−1)(−3−2)3!u3+⋯(1+u)^{-3}=1+-3 u+\frac{-3(-3-1)}{2!}u^{2}+\frac{-3(-3-1)(-3-2)}{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=−3,u=xn=-3,\quad u=x

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

  3. Substitute the small term into the series

    1+−3(x)+−3(−3−1)2!(x)2+−3(−3−1)(−3−2)3!(x)31+-3\left(x\right)+\frac{-3(-3-1)}{2!}\left(x\right)^{2}+\frac{-3(-3-1)(-3-2)}{3!}\left(x\right)^{3}

    Replace u by the actual term ready to simplify.

  4. State the constant term

    11

    The first term of (1+u)^{n} is always 1.

  5. Compute the coefficient of u

    −3-3

    The linear coefficient equals the index n.

  6. Evaluate the product n(n-1)

    −3(−3−1)=12-3(-3-1)=12

    This numerator appears in the u^{2} coefficient.

  7. Divide by 2! to get the u^{2} coefficient

    122!=6\frac{12}{2!}=6

    Dividing by 2! completes the coefficient of u^{2}.

  8. Evaluate the product n(n-1)(n-2)

    −3(−3−1)(−3−2)=−60-3(-3-1)(-3-2)=-60

    This numerator appears in the u^{3} coefficient.

  9. Divide by 3! to get the u^{3} coefficient

    −603!=−10\frac{-60}{3!}=-10

    Dividing by 3! completes the coefficient of u^{3}.

  10. Select the correct first three terms

    1−3x+6x21 - 3x + 6x^{2}

    The first three terms come from the coefficients of 1, x and x^{2}.

Answer
1−3x+6x21 - 3x + 6x^{2}
Question 5
8 markschallenging
Which of the following gives the first three terms of the expansion of (1−2x)−1\left(1-2x\right)^{-1} in ascending powers of xx?
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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=−2xn=-1,\quad u=-2x

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

  3. Substitute the small term into the series

    1+−1(−2x)+−1(−1−1)2!(−2x)2+−1(−1−1)(−1−2)3!(−2x)31+-1\left(-2x\right)+\frac{-1(-1-1)}{2!}\left(-2x\right)^{2}+\frac{-1(-1-1)(-1-2)}{3!}\left(-2x\right)^{3}

    Replace u by the actual term ready to simplify.

  4. State the constant term

    11

    The first term of (1+u)^{n} is always 1.

  5. Compute the coefficient of u

    −1-1

    The linear coefficient equals the index n.

  6. Evaluate the product n(n-1)

    −1(−1−1)=2-1(-1-1)=2

    This numerator appears in the u^{2} coefficient.

  7. Divide by 2! to get the u^{2} coefficient

    22!=1\frac{2}{2!}=1

    Dividing by 2! completes the coefficient of u^{2}.

  8. Evaluate the product n(n-1)(n-2)

    −1(−1−1)(−1−2)=−6-1(-1-1)(-1-2)=-6

    This numerator appears in the u^{3} coefficient.

  9. Divide by 3! to get the u^{3} coefficient

    −63!=−1\frac{-6}{3!}=-1

    Dividing by 3! completes the coefficient of u^{3}.

  10. Simplify the term in x

    2x2x

    Multiply the coefficient by the first power of the small term.

  11. Simplify the term in x^{2}

    4x24x^{2}

    Square the small term and multiply by its coefficient.

  12. Simplify the term in x^{3}

    8x38x^{3}

    Cube the small term and multiply by its coefficient.

  13. Write the expansion of the bracket

    1+2x+4x2+8x31 + 2x + 4x^{2} + 8x^{3}

    Collect the first four terms of (1+u)^{n}.

  14. State the range of validity

    ∣−2x∣<1 ⇒ ∣x∣<12\left|-2x\right|<1\ \Rightarrow\ \left|x\right|<\frac{1}{2}

    The series converges only for these values of x.

  15. Select the correct first three terms

    1+2x+4x21 + 2x + 4x^{2}

    The first three terms come from the coefficients of 1, x and x^{2}.

Answer
1+2x+4x21 + 2x + 4x^{2}

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