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Sequences & Series

Question 2 of 9

Binomial Theorem & Pascal's Triangle

Question

Write down the expansion of \(\left(x + \dfrac{1}{2}y\right)^{3}\)

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

    Use the Binomial Theorem to find the expansion of \(\left(x + \frac{1}{2}y\right)^3\). There will be 4 terms in the expansion.

    Use the expansion of \((a + b)^n\) where \(a = x\) and \(b = \frac{1}{2}y\) and \(n = 3\).

  2. Write out the binomial expansion
    \[ \left(x + \frac{1}{2}y\right)^{3} = \binom{3}{0}x^{3} + \binom{3}{1}(x)^{2}\left(\frac{1}{2}y\right) + \binom{3}{2}(x)\left(\frac{1}{2}y\right)^{2} + \binom{3}{3}\left(\frac{1}{2}y\right)^{3} \]
  3. Evaluate the combinations and the indices
    \[ = (1)x^{3} + (3)(x)^{2}\left(\frac{1}{2}y\right) + (3)(x)\left(\frac{1}{4}y^{2}\right) + (1)\left(\frac{1}{8}y^{3}\right) \]
  4. Evaluate the product
    \[ \left(x + \frac{1}{2}y\right)^{3} = x^{3} + \frac{3}{2}x^{2}y + \frac{3}{4}xy^{2} + \frac{1}{8}y^{3} \]

    That is, \(\left(x + \frac{1}{2}y\right)^3\) is \(x\) cubed, plus three over two \(x\) squared \(y\), plus three over four \(x\) \(y\) squared, plus one over eight \(y\) cubed.