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2007 Paper 2 Q1
D: 1600.0 B: 1516.0
generalised binomial theorem binomial expansion fractional exponent square root approximation cube root approximation numerical approximation series expansion

In this question, you are not required to justify the accuracy of the approximations.

  1. Write down the binomial expansion of \(\displaystyle \left( 1+\frac k {100} \right)^{\!\frac12}\)in ascending powers of \(k\), up to and including the \(k^3\) term.
    1. Use the value \(k=8\) to find an approximation to five decimal places for \(\sqrt{3}\,\).
    2. By choosing a suitable integer value of \(k\), find an approximation to five decimal places for \(\sqrt6\,\).
  2. By considering the first two terms of the binomial expansion of \(\displaystyle \left( 1+\frac k {1000} \right)^{\!\frac13}\), show that \(\dfrac{3029}{2100}\) is an approximation to \(\sqrt[3]{3}\).

Solution:

  1. Using the generalise binomial theorem \begin{align*} \left( 1+\frac k {100} \right)^{\frac12} &= 1 + \frac12 \frac{k}{100} + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)}{2!} \left (\frac{k}{100} \right)^2 + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)\cdot \left ( -\tfrac32\right)}{3!} \left (\frac{k}{100} \right)^3 + \cdots \\ &= 1 + \frac{1}{200}k - \frac{1}{80\,000}k^2 + \frac{1}{16\,000\,000}k^3 + \cdots \end{align*}
    1. If \(k = 8\), \begin{align*} && \left( 1+\frac 8 {100} \right)^{\frac12} &= 1 + \frac{1}{200}8 - \frac{1}{80\,000}8^2 + \frac{1}{16\,000\,000}8^3 + \cdots \\ \Rightarrow && \frac{6\sqrt{3}}{10} &\approx 1 + 0.04 - 0.0008 + 0.000032 \\ &&&= 1.039232\\ \Rightarrow && \sqrt{3} &\approx 1.73205 \, (5\, \text{d.p.}) \end{align*}
    2. If \(k = -4\), \begin{align*} && \left( 1-\frac 4 {100} \right)^{\frac12} &= 1 - \frac{1}{200}4 - \frac{1}{80\,000}4^2 - \frac{1}{16\,000\,000}4^3 + \cdots \\ \Rightarrow && \frac{4\sqrt{6}}{10} &\approx 1 -0.02-0.0002 -0.000004 \\ &&&= 0.979796\\ \Rightarrow && \sqrt{6} &\approx 2.44949\, (5\, \text{d.p.}) \end{align*}
  2. \(\,\) \begin{align*} && \left( 1+\frac k {1000} \right)^{\!\frac13} &= 1 + \frac13 \frac{k}{1000} + \cdots \\ &&&= 1 + \frac{k}{3\,000} + \cdots \\ && 3 \times 7^3 &= 1029 \\ \Rightarrow && \left( 1+\frac {29} {1000} \right)^{\!\frac13} &\approx 1 + \frac{29}{3\,000} \\ \Rightarrow && \frac{7\sqrt[3]{3}}{10} &\approx \frac{3\,029}{3000} \\ \Rightarrow && \sqrt[3]{3} &= \frac{3\,029}{2\,100} \end{align*}

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1998 Paper 2 Q2
D: 1600.0 B: 1454.6
generalised binomial theorem binomial expansion approximation square root cube root fractional exponent numerical methods

Use the first four terms of the binomial expansion of \((1-1/50)^{1/2}\), writing \(1/50 = 2/100\) to simplify the calculation, to derive the approximation \(\sqrt 2 \approx 1.414214\). Calculate similarly an approximation to the cube root of 2 to six decimal places by considering \((1+N/125)^a\), where \(a\) and \(N\) are suitable numbers. [You need not justify the accuracy of your approximations.]

Solution: \begin{align*} && (1-1/50)^{1/2} &= 1 + \frac12 \cdot \left ( -\frac1{50} \right) + \frac1{2!} \frac12 \cdot \left ( -\frac12 \right)\cdot \left ( -\frac1{50} \right)^2 + \frac1{3!} \frac12 \cdot \left ( -\frac12 \right) \cdot \left ( -\frac32 \right)\cdot \left ( -\frac1{50} \right)^3 + \cdots \\ &&&=1-\frac{1}{100} - \frac12 \frac1{10000} -\frac12 \frac1{1000000} +\cdots \\ &&&= 0.9899495 + \cdots \\ \Rightarrow && \frac{7\sqrt{2}}{10} &\approx 0.9899495 \\ \Rightarrow && \sqrt{2} &\approx \frac{9.899495}{7} \\ &&&\approx 1.414214 \end{align*} \begin{align*} && (1 + 3/125)^{1/3} &= \frac{\sqrt[3]{125+3}}{5} \\ &&& = \frac{8\sqrt[3]{2}}{10} \\ && (1 + 3/125)^{1/3} &= 1 + \frac13 \left ( \frac{3}{125} \right) + \frac1{2!} \cdot \frac{1}{3} \cdot \left ( -\frac23\right) \left ( \frac{3}{125}\right)^2 +\cdots \\ &&&= 1+ \frac{8}{1000} - \frac{64}{1000000} \\ &&&= 1.007936 \\ \Rightarrow && \sqrt[3]{2} &= \frac{10.07936}{8} \\ &&&= 1.259920 \end{align*}

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