Standard High contrast

Sequences & Series

Question 3 of 9

Binomial Theorem & Pascal's Triangle

Question

Find the term in \(x^3\) of the expansion \(\left(\dfrac{1}{2}x - 4\right)^{4}\)

Show solution Hide solution Fully worked — 4 steps
  1. Set up the problem

    Use the Binomial Theorem to find the term that will give \(x^3\) in the expansion of \(\left(\frac{1}{2}x - 4\right)^4\).

    We only want to find the term in \(x^3\) so we don't need the complete expansion.

    Let \(a = \frac{1}{2}x\), \(b = -4\), \(n = 4\), \(n - k = 3\) (since the index of \(x\) in the first term is 1).

  2. To solve for \(k\)

    Substitute \(n = 4\) into \(n - k = 3\):

    \[ 4 - k = 3 \]

    Subtract 4 from both sides:

    \[ -k = 3 - 4 \]

    Subtract 4 from 3, then multiply both sides by \(-1\):

    \[ k = 1 \]
  3. Substitute into the general term

    Substitute \(a\), \(b\), \(n\), \(k\), \(n - k\) into \(\binom{n}{k}a^{n-k}b^{k}\) to get the term with \(x^3\):

    \[ \binom{n}{k}a^{n-k}b^{k} = \binom{4}{1}\left(\frac{1}{2}x\right)^{3}(-4) \]

    Evaluate \(\binom{4}{1}\) and \(\left(\frac{1}{2}x\right)^{3}\):

    \[ = (4)\left(\frac{1}{8}x^{3}\right)(-4) \]

    Evaluate the product:

    \[ = -2x^{3} \]
  4. State the term

    Therefore the term in \(x^3\) in the expansion is \(-2x^3\): minus 2 \(x\) cubed.