Binomial Theorem
Properties of Binomial Coefficients
Grade 11
Question:
<p>Given \(\sum_{r=0}^{2n} a_r \left(\frac{1}{x}\right)^r = \left(1 + \frac{1}{x} + \frac{1}{x^2}\right)^n\), find \(a_r\) in terms of \(a_{2n-r}\) for \(0 \le r \le 2n\).</p>
<p>\(a_r = a_{n-r}\)</p>
<p>\(a_r = a_{2n+r}\)</p>
<p>\(a_r = a_{2n-r}\)</p>
<p>\(a_r = a_{r-n}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the right side is symmetric in its expansion, and comparing coefficients of x^r on both sides reveals a reciprocal relationship through the structure (1 + 1/x + 1/x²)^n = x^(-2n)(x² + x + 1)^n.
<p><strong>Step 1:</strong> Expand the right side using the binomial theorem:</p><p>$$\left(1 + \frac{1}{x} + \frac{1}{x^2}\right)^n = \sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{x} + \frac{1}{x^2}\right)^k$$</p><p><strong>Step 2:</strong> For each k, expand $\left(\frac{1}{x} + \frac{1}{x^2}\right)^k = \sum_{j=0}^{k}\binom{k}{j}\frac{1}{x^{k+j}}$</p><p>The coefficient of $\frac{1}{x^r}$ comes from all combinations where $k + j = r$, giving:</p><p>$$a_r = \sum_{k=0}^{\min(r,n)}\binom{n}{k}\binom{k}{r-k}$$</p><p><strong>Step 3:</strong> Replace r with (2n-r):</p><p>$$a_{2n-r} = \sum_{k=0}^{\min(2n-r,n)}\binom{n}{k}\binom{k}{2n-2k-r}$$</p><p><strong>Step 4:</strong> By the symmetry of binomial coefficients and the constraint that indices must be non-negative, the valid range of k ensures that:</p><p>$$\boxed{a_r = a_{2n-r}}$$</p><p>This means the coefficients are symmetric about the middle term. ∴ Answer: C</p>
Correct Answer: C