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Sequences & Series

Question 9 of 9

Binomial Theorem & Pascal's Triangle

Question

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

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

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

    We only want to find the coefficient of the term in \(\frac{1}{x^2}\) so we don't need the complete expansion. Let us examine the indices of the \(x\)-variable.

  2. Determine the powers of each term

    If we have \((x^2)^3\left(\frac{1}{x^3}\right)\) we will get \(x^3\). To get \(\frac{1}{x^2}\), we have to raise \(x^2\) to the 2nd power, which means we have to raise \(\frac{1}{x^3}\) to the 2nd power since the indices must sum up to 4.

    So, we let \(a = \frac{1}{4}x^2\), \(b = \frac{1}{4x^3}\), \(n = 4\), \(n - k = 2\) and \(k = 2\).

  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 \(\frac{1}{x^2}\):

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

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

    \[ = (6)\left(\frac{1}{16}x^{4}\right)\left(\frac{1}{16x^{6}}\right) \]

    Evaluate the product:

    \[ = \frac{3}{128x^{2}} \]
  4. State the coefficient

    Therefore, the coefficient of the term in \(\frac{1}{x^2}\) in the expansion is \(\frac{3}{128}\): 3 over 128.