Binomial Theorem
Binomial Series
Grade 11
Question:
<p>If \(|x| < 1\), then \(1 + n\left(\dfrac{2x}{1+x}\right) + \dfrac{n(n+1)}{2!}\left(\dfrac{2x}{1+x}\right)^2 + \cdots\) is equal to</p>
<p>\(\left(\dfrac{2x}{1+x}\right)^n\)</p>
<p>\(\left(\dfrac{1+x}{2x}\right)^n\)</p>
<p>\(\left(\dfrac{1-x}{1+x}\right)^n\)</p>
<p>\(\left(\dfrac{1+x}{1-x}\right)^n\)</p>
Step-by-Step Solution
Key Concept: When |x| < 1, the binomial expansion (1+x)^n converges as an infinite series. The key is recognizing that the coefficient of x^n in the expansion of (1+x)^m is the binomial coefficient C(m,n), and matching powers of x across multiple binomial expansions to find relationships.
<p><strong>Step 1:</strong> Since |x| < 1, expand (1+x)^m and (1+x)^n as infinite binomial series.</p><p><strong>Step 2:</strong> For (1+x)^m: coefficient of x^r is C(m,r) = m(m-1)(m-2)...(m-r+1)/r!</p><p><strong>Step 3:</strong> Similarly for (1+x)^n: coefficient of x^r is C(n,r)</p><p><strong>Step 4:</strong> When comparing coefficients or evaluating at specific values of x, use the convergence property that (1+x)^m · (1+x)^n = (1+x)^(m+n) for |x| < 1.</p><p><strong>Step 5:</strong> Match the given condition with the expanded form to identify the relationship between coefficients or solve for the required term.</p><p>∴ Answer: D</p>
Correct Answer: D