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Sequences & Series

Question 4 of 9

Binomial Theorem & Pascal's Triangle

Question

Find the term in \(x^3\) of the expansion \(\left(3x^3 - \dfrac{1}{2}\right)^{5}\)

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 \(\left(3x^3 - \frac{1}{2}\right)^5\).

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

    Let \(a = 3x^3\), \(b = -\frac{1}{2}\), \(n = 5\), \(n - k = 1\) (Since \(x\) in the 1st term is already in \(x^3\), we just need to raise it to the 1st power to maintain \(x^3\).)

  2. To solve for \(k\)

    Substitute \(n = 5\) into \(n - k = 1\):

    \[ 5 - k = 1 \]

    Subtract 5 from both sides:

    \[ -k = 1 - 5 \]

    Subtract 5 from 1, then multiply both sides by \(-1\):

    \[ k = 4 \]
  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 \(x^3\):

    \[ \binom{n}{k}a^{n-k}b^{k} = \binom{5}{4}\left(3x^{3}\right)^{1}\left(-\frac{1}{2}\right)^{4} \]

    Evaluate \(\binom{5}{4}\) and \(\left(-\frac{1}{2}\right)^{4}\):

    \[ = (5)(3x^{3})\left(\frac{1}{16}\right) \]

    Evaluate the product:

    \[ = \frac{15}{16}x^{3} \]
  4. State the term

    Therefore the term in \(x^3\) in the expansion is \(\frac{15}{16}x^{3}\): 15 over 16 \(x\) cubed.