Binomial Theorem
Coefficient comparison
Grade 11
Question:
<p>If \(\displaystyle\sum_{r=0}^{2n} a_r(x-2)^r = \sum_{r=0}^{2n} b_r(x-3)^r\) and \(a_k = 1\) for all \(k \ge n\), then \(b_n\) is equal to</p>
<p>(1) \({}^{2n+1}C_{n-1}\)</p>
<p>(2) \({}^{2n}C_{n+1}\)</p>
<p>(3) \({}^{2n}C_n\)</p>
<p>(4) \({}^{2n+1}C_{n+1}\)</p>
Step-by-Step Solution
Key Concept: Both sides represent the same polynomial in different forms. Since $a_k = 1$ for $k \geq n$, the left side has $(x-2)^n + (x-2)^{n+1} + \cdots + (x-2)^{2n}$ as its tail. Use substitution $x = 3$ strategically and expand using binomial theorem to find $b_n$ as the coefficient of $(x-3)^n$.
<p><strong>Step 1:</strong> Recognize that for $k \geq n$, we have $a_k = 1$. Write the left side as:</p><p>$$\sum_{r=0}^{n-1} a_r(x-2)^r + \sum_{r=n}^{2n} (x-2)^r$$</p><p><strong>Step 2:</strong> The geometric series gives: $\sum_{r=n}^{2n} (x-2)^r = (x-2)^n \cdot \frac{(x-2)^{n+1}-1}{(x-2)-1} = (x-2)^n \cdot \frac{(x-2)^{n+1}-1}{x-3}$</p><p><strong>Step 3:</strong> Since both sides equal the same polynomial, they're equal for all $x$. The coefficient of $(x-3)^n$ on the right is $b_n$. On the left, the term $(x-2)^n$ contributes to $b_n$ when expanded in powers of $(x-3)$.</p><p><strong>Step 4:</strong> Substitute $x = 3$ in the derivative relation or use: $(x-2)^n = [(x-3)+1]^n$. The coefficient of $(x-3)^n$ in $(x-2)^n$ is $\binom{n}{n} = 1$.</p><p><strong>Step 5:</strong> For the geometric series tail: $\sum_{r=n}^{2n}(x-2)^r$ evaluated at coefficient of $(x-3)^n$ requires binomial expansion. $(x-2)^r = [(x-3)+1]^r$ gives coefficient $\binom{r}{n}$ for $(x-3)^n$.</p><p><strong>Step 6:</strong> Thus $b_n = \sum_{r=n}^{2n}\binom{r}{n} = \binom{n}{n} + \binom{n+1}{n} + \cdots + \binom{2n}{n} = \binom{2n+1}{n+1}$</p><p>∴ Answer: D</p>
Correct Answer: D