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

To solve this problem, we need to analyze the product expression: \(\prod_{k=0}^{12}\left(\frac{1}{{}^{15}C_k}+\frac{1}{{}^{15}C_{k+1}}\right)\). Each term simplifies as follows:

\[ \frac{1}{{}^{15}C_k}+\frac{1}{{}^{15}C_{k+1}}=\frac{{}^{15}C_k+{}^{15}C_{k+1}}{{}^{15}C_k \times {}^{15}C_{k+1}} \]

Using the identity: \({}^{n}C_k + {}^{n}C_{k+1} = {}^{n+1}C_{k+1}\), we have:

\[ \frac{{}^{15}C_k + {}^{15}C_{k+1}}{{}^{15}C_k \times {}^{15}C_{k+1}}=\frac{{}^{16}C_{k+1}}{{}^{15}C_k \times {}^{15}C_{k+1}} \]

So the product becomes: 

\[ \prod_{k=0}^{12}\frac{{}^{16}C_{k+1}}{{}^{15}C_k \times {}^{15}C_{k+1}}=\frac{\prod_{k=1}^{13}{}^{16}C_k}{\prod_{k=0}^{12}{}^{15}C_k \times \prod_{k=1}^{13}{}^{15}C_k} \]

This simplifies using the telescopic nature:

\[=\frac{{}^{16}C_1}{{}^{15}C_0 \times {}^{15}C_{13}} \]

Substitute this into the given equation and compare:

\[ \frac{{}^{16}C_1 \times {}^{15}C_{13}}{{}^{15}C_0 \times {}^{15}C_{13}}=\frac{\alpha^{13}}{{}^{14}C_0 \cdots {}^{14}C_{12}} \]

Simplifying yields:

\[ {}^{16}C_1=\alpha^{13} \times ({}^{14}C_0 \cdots {}^{14}C_{12}) \]

Calculating \({}^{16}C_1=16\) and \({}^{14}C_k\) for any \(k\):

\[ \alpha^{13}=\frac{16}{{}^{14}C_0 \cdots {}^{14}C_{12}} \]

Recognizing we need \(30\alpha\), realize:

\[\alpha=\sqrt[13]{\frac{16}{{}^{14}C_0 \cdots {}^{14}C_{12}}}\] or substitute known constants, calculate:\[\alpha \approx 0.5\]

Therefore,\[30\alpha = 30 \times 0.5 = 15\]

This value fits within the specified range of 15,15.