Permutations & Combinations
Binomial Coefficients and Products
Grade 11
Question:
<p>Here, \(a_r = {}^{2n}C_r\). Find the value of \(\displaystyle\prod_{r=1}^{2n}\left(1+\dfrac{a_r}{a_{r-1}}\right)\).</p>
<p>\(\dfrac{(2n+1)^{2n}}{(2n)!}\)</p>
<p>\(\dfrac{(n+1)^{2n}}{(2n)!}\)</p>
<p>\(\dfrac{(2n+1)^n}{n!}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that $1 + \frac{a_r}{a_{r-1}} = 1 + \frac{\binom{2n}{r}}{\binom{2n}{r-1}} = \frac{\binom{2n}{r-1} + \binom{2n}{r}}{\binom{2n}{r-1}} = \frac{\binom{2n+1}{r}}{\binom{2n}{r-1}}$ using the Pascal's triangle identity, which creates a telescoping product.
<p><strong>Step 1:</strong> Simplify each factor using the ratio of consecutive binomial coefficients:</p><p>$$\frac{a_r}{a_{r-1}} = \frac{\binom{2n}{r}}{\binom{2n}{r-1}} = \frac{(2n)!/(r!(2n-r)!)}{(2n)!/((r-1)!(2n-r+1)!)} = \frac{(r-1)!(2n-r+1)!}{r!(2n-r)!} = \frac{2n-r+1}{r}$$</p><p><strong>Step 2:</strong> Therefore:</p><p>$$1 + \frac{a_r}{a_{r-1}} = 1 + \frac{2n-r+1}{r} = \frac{r + 2n - r + 1}{r} = \frac{2n+1}{r}$$</p><p><strong>Step 3:</strong> The product becomes:</p><p>$$\prod_{r=1}^{2n}\frac{2n+1}{r} = (2n+1)^{2n} \cdot \prod_{r=1}^{2n}\frac{1}{r} = (2n+1)^{2n} \cdot \frac{1}{(2n)!}$$</p><p><strong>Step 4:</strong> Alternatively, rewrite as:</p><p>$$\prod_{r=1}^{2n}\frac{2n+1}{r} = \frac{(2n+1)^{2n}}{(2n)!}$$</p><p>∴ Answer: $\dfrac{(2n+1)^{2n}}{(2n)!}$ or equivalently $(2n+1)^{2n}/(2n)!$</p>
Correct Answer: A