Photo AI
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 2 - 2010 - Paper 7
Question 2
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. (b) 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 2 - 2010 - Paper 7
Step 1
Find the binomial expansion of \(\sqrt{1-8x}\)
Answer
To expand (\sqrt{1-8x}), we use the binomial expansion formula ((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots), where (u = -8x) and (n = \frac{1}{2}).
-
Substitute values:
[\sqrt{1-8x} = (1 + (-8x))^{\frac{1}{2}} ] This gives us: [= 1 + \frac{1}{2}(-8x) + \frac{\frac{1}{2}(-\frac{1}{2})}{2!}(-8x)^2 + \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{3!}(-8x)^3 + \ldots ] -
Calculate the terms:
- First term: (1)
- Second term: (-4x)
- Third term: (\frac{32x^2}{2} = 16x^2)
- Fourth term: (-\frac{32 \cdot 8x^3}{6} = -\frac{128x^3}{6} = -\frac{64x^3}{3})
Thus, combining them: [\sqrt{1-8x} \approx 1 - 4x + 16x^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}):
-
Calculate the expression:
[\sqrt{1 - 8 \times \frac{1}{100}} = \sqrt{1 - \frac{8}{100}} = \sqrt{\frac{92}{100}} = \frac{\sqrt{92}}{10}] -
Simplify further:
[\sqrt{92} = \sqrt{4 \times 23} = 2\sqrt{23}]
Therefore:
[\frac{\sqrt{92}}{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
Using the expansion from part (a):
[\sqrt{1-8x} \approx 1 - 4\left(\frac{1}{100}\right) + 16\left(\frac{1}{100}\right)^2 - \frac{64}{3}\left(\frac{1}{100}\right)^3]
-
Calculate each term:
- First term: (1)
- Second term: (-0.04)
- Third term: (0.016)
- Fourth term: (-0.000064)
-
Combine the terms:
[= 1 - 0.04 + 0.016 - 0.000064 \approx 0.959168] -
Approximate (\sqrt{23}):
Since (\sqrt{23} = 5\times0.959168), we find:
[\sqrt{23} \approx 4.79584]
Thus, the final answer to 5 decimal places is (4.79584).