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Find the binomial expansion of \[\sqrt{(1 - 8x)}, \ |x| < \frac{1}{8},\] in ascending powers of x up to and including the term in \(x^3\), simplifying each term - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 7
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Find the binomial expansion of \[\sqrt{(1 - 8x)}, \ |x| < \frac{1}{8},\] in ascending powers of x up to and including the term in \(x^3\), simplifying each term. Sh... show full transcript
Worked Solution & Example Answer:Find the binomial expansion of \[\sqrt{(1 - 8x)}, \ |x| < \frac{1}{8},\] in ascending powers of x up to and including the term in \(x^3\), simplifying each term - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 7
Step 1
Find the binomial expansion of \(\sqrt{(1 - 8x)}\)
Answer
To find the binomial expansion of (\sqrt{(1 - 8x)}), we use the binomial series: ( (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + ... ). For (n = \frac{1}{2}) and (u = -8x), we have:
[ \sqrt{(1 - 8x)} = (1 - 8x)^{1/2} = 1 - \frac{8x}{2} - \frac{(8x)^2}{2 \cdot 2} + \frac{(8x)^3}{2 \cdot 3!} + ... ] [ = 1 - 4x - 32x^2 + \frac{128x^3}{6} + ... ] [ = 1 - 4x - 32x^2 + \frac{64}{3}x^3 + ... ]
Step 2
Show that, when \(x = \frac{1}{100}\), the exact value of \(\sqrt{(1 - 8x)}\) is \(\frac{\sqrt{23}}{5}\)
Answer
Substituting (x = \frac{1}{100}) into (1 - 8x): [ 1 - 8 \left(\frac{1}{100}\right) = 1 - \frac{8}{100} = 1 - 0.08 = 0.92. ]
Thus, (\sqrt{(1 - 8x)} = \sqrt{0.92} = \frac{\sqrt{92}}{10} = \frac{\sqrt{23 \cdot 4}}{10} = \frac{2\sqrt{23}}{10} = \frac{\sqrt{23}}{5}.]
Step 3
Substitute \(x = \frac{1}{100}\) into the binomial expansion in part (a) and hence obtain an approximation to \(\sqrt{23}\)
Answer
Now, substitute (x = \frac{1}{100}) into the binomial expansion derived in part (a): [ 1 - 4\left(\frac{1}{100}\right) - 32\left(\frac{1}{100}\right)^2 + \frac{64}{3}\left(\frac{1}{100}\right)^3. ]
Calculating each term: [ = 1 - 0.04 - 0.00032 + \frac{64}{3 \cdot 100^3}. ]
Estimating (\sqrt{23}): [ \sqrt{23} \approx 5 \cdot (1 - 0.04 - 0.00032) = 5 \cdot 0.959168 = 4.79584. ] Thus, the approximation to five decimal places is (4.79584).