Hard A-Level Binomial expansion (rational powers) Questions

Challenging, exam-style A-Level Binomial expansion (rational powers) questions with worked solutions. Stretch yourself on the hardest binomial expansion, rational index, partial fractions problems.

binomial expansionrational indexpartial fractions
A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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}
Question 2
8 markschallenging
In the expansion of (1−2x)−1\left(1-2x\right)^{-1}, what is the coefficient of x3x^{3}?
Show worked solution

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 coefficient

    coefficient of x3=8\text{coefficient of }x^{3}=8

    The coefficient of x^{3} is read from the expansion.

Answer
88
Question 3
8 markschallenging
For which values of xx is the expansion of (16+x)12\left(16+x\right)^{\frac{1}{2}} valid?
Show worked solution

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=116xn=\frac{1}{2},\quad u=\frac{1}{16}x

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

  3. Take out a factor of 16

    (16+x)12=1612(1+116x)12(16+x)^{\frac{1}{2}}=16^{\frac{1}{2}}\left(1+\frac{1}{16}x\right)^{\frac{1}{2}}

    Writing it as a^{n}(1+u)^{n} lets us apply the standard series.

  4. Evaluate the constant factor 16^{\frac{1}{2}}

    1612=416^{\frac{1}{2}}=4

    This constant multiplies every term of the expansion.

  5. Substitute the small term into the series

    1+12(116x)+12(12−1)2!(116x)2+12(12−1)(12−2)3!(116x)31+\frac{1}{2}\left(\frac{1}{16}x\right)+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}\left(\frac{1}{16}x\right)^{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}\left(\frac{1}{16}x\right)^{3}

    Replace u by the actual term ready to simplify.

  6. State the constant term

    11

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

  7. Compute the coefficient of u

    12\frac{1}{2}

    The linear coefficient equals the index n.

  8. Evaluate the product n(n-1)

    12(12−1)=−14\frac{1}{2}(\frac{1}{2}-1)=- \frac{1}{4}

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

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

    −142!=−18\frac{- \frac{1}{4}}{2!}=- \frac{1}{8}

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

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

    12(12−1)(12−2)=38\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)=\frac{3}{8}

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

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

    383!=116\frac{\frac{3}{8}}{3!}=\frac{1}{16}

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

  12. Simplify the term in x

    132x\frac{1}{32}x

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

  13. Simplify the term in x^{2}

    −12048x2- \frac{1}{2048}x^{2}

    Square the small term and multiply by its coefficient.

  14. Simplify the term in x^{3}

    165536x3\frac{1}{65536}x^{3}

    Cube the small term and multiply by its coefficient.

  15. Select the correct range of validity

    ∣x∣<16\left|x\right|<16

    The expansion is valid only when the small term has modulus below 1.

Answer
∣x∣<16\left|x\right|<16
Question 4
8 markschallenging
Which of the following gives the first three terms of the expansion of (1+3x)−1\left(1+3x\right)^{-1} in ascending powers of xx?
Show worked solution

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=3xn=-1,\quad u=3x

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

  3. Substitute the small term into the series

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

    −3x-3x

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

  11. Simplify the term in x^{2}

    9x29x^{2}

    Square the small term and multiply by its coefficient.

  12. Simplify the term in x^{3}

    −27x3-27x^{3}

    Cube the small term and multiply by its coefficient.

  13. Write the expansion of the bracket

    1−3x+9x2−27x31 - 3x + 9x^{2} - 27x^{3}

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

  14. State the range of validity

    ∣3x∣<1 ⇒ ∣x∣<13\left|3x\right|<1\ \Rightarrow\ \left|x\right|<\frac{1}{3}

    The series converges only for these values of x.

  15. Select the correct first three terms

    1−3x+9x21 - 3x + 9x^{2}

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

Answer
1−3x+9x21 - 3x + 9x^{2}
Question 5
8 markschallenging
In the expansion of (1+x)−2\left(1+x\right)^{-2}, what is the coefficient of x3x^{3}?
Show worked solution

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. Evaluate the product n(n-1)

    −2(−2−1)=6-2(-2-1)=6

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

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

    62!=3\frac{6}{2!}=3

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

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

    −2(−2−1)(−2−2)=−24-2(-2-1)(-2-2)=-24

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

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

    −243!=−4\frac{-24}{3!}=-4

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

  10. Simplify the term in x

    −2x-2x

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

  11. Simplify the term in x^{2}

    3x23x^{2}

    Square the small term and multiply by its coefficient.

  12. Simplify the term in x^{3}

    −4x3-4x^{3}

    Cube the small term and multiply by its coefficient.

  13. Write the expansion of the bracket

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

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

  14. State the range of validity

    ∣x∣<1 ⇒ ∣x∣<1\left|x\right|<1\ \Rightarrow\ \left|x\right|<1

    The series converges only for these values of x.

  15. Select the correct coefficient

    coefficient of x3=−4\text{coefficient of }x^{3}=-4

    The coefficient of x^{3} is read from the expansion.

Answer
−4-4

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