Question

\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]

Then \[ 30\alpha = \underline{\hspace{1cm}} \] 

Hint:

Identities like \( \frac{1}{{}^{n}C_r} + \frac{1}{{}^{n}C_{r+1}} = \frac{n+1}{n \cdot {}^{n-1}C_r} \) are extremely useful for simplifying series or products involving reciprocals of binomial coefficients.

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
The question involves products of reciprocals of binomial coefficients. We can use the identity \( \frac{1}{{}^{n}C_r} + \frac{1}{{}^{n}C_{r+1}} = \frac{n+1}{n \cdot {}^{n-1}C_r} \) (derived from basic factorial definitions) to simplify each term.

Step 2: Key Formula or Approach:
Consider the general term: \( T_r = \frac{1}{{}^{n}C_r} + \frac{1}{{}^{n}C_{r+1}} \).
Using \( {}^{n}C_{r+1} = \frac{n}{r+1} {}^{n-1}C_r \):
\[ T_r = \frac{r!(n-r)!}{n!} + \frac{(r+1)!(n-r-1)!}{n!} = \frac{r!(n-r-1)!}{n!} [ (n-r) + (r+1) ] = \frac{r!(n-r-1)!}{n!} (n+1) \]
\[ T_r = \frac{n+1}{n} \cdot \frac{r!(n-r-1)!}{(n-1)!} = \frac{n+1}{n \cdot {}^{n-1}C_r} \]

Step 3: Detailed Explanation:
Applying this formula with \( n = 15 \):
\[ \frac{1}{{}^{15}C_r} + \frac{1}{{}^{15}C_{r+1}} = \frac{15+1}{15 \cdot {}^{14}C_r} = \frac{16}{15 \cdot {}^{14}C_r} \]
The given product is:
\[ \prod_{r=0}^{12} \left( \frac{1}{{}^{15}C_r} + \frac{1}{{}^{15}C_{r+1}} \right) = \prod_{r=0}^{12} \frac{16}{15 \cdot {}^{14}C_r} \]
Since the product has 13 terms (from \( r=0 \) to \( r=12 \)):
\[ \text{Product} = \frac{(16/15)^{13}}{\prod_{r=0}^{12} {}^{14}C_r} = \frac{(16/15)^{13}}{{}^{14}C_0 {}^{14}C_1 \dots {}^{14}C_{12}} \]
Comparing this with the given form \( \frac{\alpha^{13}}{{}^{14}C_0 {}^{14}C_1 \dots {}^{14}C_{12}} \), we get:
\[ \alpha = \frac{16}{15} \]
We need to find \( 30\alpha \):
\[ 30\alpha = 30 \times \frac{16}{15} = 2 \times 16 = 32 \]

Step 4: Final Answer:
The value of \( 30\alpha \) is 32.