<p>If \(|x| < 1\), then the coefficient of \(x^n\) in expansion of \((1 + x + x^2 + x^3 + \cdots)^2\) is</p>
Step-by-Step Solution
Key Concept: Use the binomial series expansion for fractional powers: $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + ...$, valid when $|u| < 1$. Here, identify $u = -x^2$ and $n = -1/2$ to expand $(1-x^2)^{-1/2}$.
<p><strong>Step 1:</strong> Rewrite the expression as $(1-x^2)^{-1/2}$ and apply the binomial series formula $(1+u)^n = \sum_{r=0}^{\infty} \binom{n}{r}u^r$ with $u = -x^2$ and $n = -\frac{1}{2}$.</p><p><strong>Step 2:</strong> Calculate successive binomial coefficients:</p><ul><li>$\binom{-1/2}{0} = 1$</li><li>$\binom{-1/2}{1} = -\frac{1}{2}$, so term is $-\frac{1}{2}(-x^2) = \frac{1}{2}x^2$</li><li>$\binom{-1/2}{2} = \frac{(-1/2)(-3/2)}{2!} = \frac{3}{8}$, so term is $\frac{3}{8}x^4$</li><li>$\binom{-1/2}{3} = \frac{(-1/2)(-3/2)(-5/2)}{3!} = -\frac{5}{16}$, so term is $-\frac{5}{16}x^6$</li></ul><p><strong>Step 3:</strong> Combine to get: $(1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \frac{5}{16}x^6 + ...$</p><p>∴ Answer: D</p>
Correct Answer: D