Binomial Theorem
Binomial Series
Grade 11
Question:
<p>The coefficient of \(x^4\) in the expansion of \(\left(\sqrt{1+x^2} - x\right)^{-1}\) in ascending powers of \(x\), when \(|x| < 1\), is</p>
<p>0</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(-\dfrac{1}{2}\)</p>
<p>\(-\dfrac{1}{8}\)</p>
Step-by-Step Solution
Key Concept: Rewrite the expression as a reciprocal of a binomial, then use the binomial series expansion (1+u)^(-1) = 1 - u + u² - u³ + ... for |u| < 1, where u = x - √(1+x²).
<p><strong>Step 1:</strong> Rationalize by multiplying numerator and denominator:</p><p>$$\left(\sqrt{1+x^2} - x\right)^{-1} = \frac{\sqrt{1+x^2} + x}{(\sqrt{1+x^2})^2 - x^2} = \frac{\sqrt{1+x^2} + x}{1}$$</p><p><strong>Step 2:</strong> Expand √(1+x²) using binomial series (1+u)^(1/2) with u = x²:</p><p>$$\sqrt{1+x^2} = 1 + \frac{1}{2}x^2 + \frac{(1/2)(-1/2)}{2!}x^4 + ... = 1 + \frac{x^2}{2} - \frac{x^4}{8} + ...$$</p><p><strong>Step 3:</strong> Add x to get the full expansion:</p><p>$$(\sqrt{1+x^2} + x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} + ...$$</p><p><strong>Step 4:</strong> The coefficient of x⁴ comes from the explicit term and the product of lower-order terms. From the expansion: coefficient of x⁴ = <strong>-1/8</strong></p><p>∴ Answer: A</p>
Correct Answer: A