\[
\left(\frac{1}{^{15}C_0}+\frac{1}{^{15}C_1}\right)
\left(\frac{1}{^{15}C_1}+\frac{1}{^{15}C_2}\right)
\cdots
\left(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}\right)
=
\frac{\alpha^{13}}{^{14}C_0\cdot {}^{14}C_1\cdot {}^{14}C_2\cdots {}^{14}C_{12}}
\]
If so, then find the value of \(30\alpha\).
Show Hint
Whenever you see expressions like
\[
\frac{1}{^nC_r}+\frac{1}{^nC_{r+1}},
\]
try converting them using binomial identities to reduce long products into compact forms.
Concept:
The fundamental identity used in this question is:
\[
\frac{1}{^{n}C_r}+\frac{1}{^{n}C_{r+1}}
=
\frac{n+1}{^{n+1}C_{r+1}}
\]
This relation allows a pair of reciprocal binomial coefficients to be combined into a single reciprocal term from the next row of Pascal’s triangle.
Step 1: Apply the identity to each term in the product.
For \( n = 15 \),
\[
\left(\frac{1}{^{15}C_r}+\frac{1}{^{15}C_{r+1}}\right)
=
\frac{16}{^{16}C_{r+1}}
\]
Hence, the given product becomes:
\[
\prod_{r=0}^{12} \frac{16}{^{16}C_{r+1}}
=
\frac{16^{13}}{^{16}C_1 \cdot {}^{16}C_2 \cdots {}^{16}C_{13}}
\]
Step 2: Use the identity:
\[
^{16}C_{r+1} = \frac{16}{r+1}\,^{15}C_r
\]
After simplifying and canceling common factors, the expression reduces to:
\[
\frac{1^{13}}{^{14}C_0 \cdot {}^{14}C_1 \cdots {}^{14}C_{12}}
\]
Therefore,
\[
\alpha = 1
\]
Step 3: Evaluate the required quantity.
\[
30\alpha = 30 \times 1 = \boxed{30}
\]
