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Sequences & Series

Question 7 of 9

Binomial Theorem & Pascal's Triangle

Question

Find the coefficient of the term \(x^6\) of the expansion \(\left(\dfrac{1}{4}x^2 + 5\right)^{7}\)

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

    Use the binomial theorem to find the term that will give \(x^6\) in the expansion of \(\left(\frac{1}{4}x^2 + 5\right)^7\).

    We only want to find the coefficient of the term in \(x^6\) so we don't need the complete expansion.

    Let \(a = \frac{1}{4}x^2\), \(b = 5\), \(n = 7\), \(n - k = 3\) (since the index of \(x\) in the first term is already 2).

  2. To solve for \(k\)

    Substitute \(n = 7\) into \(n - k = 3\):

    \[ 7 - k = 3 \]

    Subtract 7 from both sides:

    \[ -k = 3 - 7 \]

    Subtract 7 from 3, 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 \(x^6\):

    \[ \binom{n}{k}a^{n-k}b^{k} = \binom{7}{4}\left(\frac{1}{4}x^{2}\right)^{3}(5)^{4} \]

    Evaluate \(\binom{7}{4}\), \(\left(\frac{1}{4}x^{2}\right)^{3}\) and \((5)^{4}\):

    \[ = (35)\left(\frac{1}{64}x^{6}\right)(625) \]

    Evaluate the product:

    \[ = \frac{21875}{64}x^{6} \]
  4. State the coefficient

    Therefore, the coefficient of the term in \(x^6\) in the expansion is \(\frac{21875}{64}\): 21875 over 64.