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Question 1 of 9

Binomial Theorem & Pascal's Triangle

Question

If \(x\) is too small so that terms of \(x^3\) and higher can be ignored, show that:

\[ \left(\frac{1}{2} + x\right)(1 + x)^{4} = \frac{1}{2} + 3x + 7x^{2}. \]
Show solution Hide solution Fully worked — 4 steps
  1. Expand \((1+x)^4\) using Pascal's Triangle

    We use Pascal's Triangle in the expansion of \((1 + x)^4\). The index of \((1 + x)^4\) is 4, so we look on the 5th line of Pascal's Triangle and the coefficients are 1, 4, 6, 4 and 1.

    \[ \left(\frac{1}{2} + x\right)(1 + x)^{4} = \left(\frac{1}{2} + x\right)\left(1 + 4x + 6x^{2} + 4x^{3} + x^{4}\right) \]
  2. Multiply out the brackets
    \[ = \frac{1}{2} + 2x + 3x^{2} + 2x^{3} + \frac{1}{2}x^{4} + x + 4x^{2} + 6x^{3} + 4x^{4} + x^{5} \]
  3. Simplify the product
    \[ = \frac{1}{2} + 3x + 7x^{2} + 8x^{3} + \frac{9}{2}x^{4} + x^{5} \]
  4. Omit the term containing \(x^3\) and higher

    Omit the term containing \(x^3\) and higher since \(x\) is too small so that terms of \(x^3\) and higher can be ignored:

    \[ \left(\frac{1}{2} + x\right)(1 + x)^{4} = \frac{1}{2} + 3x + 7x^{2} \]

    That is, ignoring these terms, \(\left(\frac{1}{2} + x\right)(1 + x)^4\) is one half plus 3 \(x\) plus 7 \(x\) squared, as required.