Question

\[ \left(\frac{1}{^{15}C_0}+\frac{1}{^{15}C_1}\right) \left(\frac{1}{^{15}C_1}+\frac{1}{^{15}C_2}\right) \cdots \left(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}\right) = \frac{\alpha^{13}}{^{14}C_0\cdot {}^{14}C_1\cdot {}^{14}C_2\cdots {}^{14}C_{12}} \] If so, then find the value of \(30\alpha\).

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.

Answer 1

Correct Answer: 30

Concept: The key identity used in this problem is: \[ \frac{1}{^{n}C_r}+\frac{1}{^{n}C_{r+1}} = \frac{n+1}{^{n+1}C_{r+1}} \] This identity helps convert sums of reciprocals of binomial coefficients into a single reciprocal term of the next row of Pascal’s triangle.
Step 1: Apply the identity to each bracket. For \(n=15\), \[ \left(\frac{1}{^{15}C_r}+\frac{1}{^{15}C_{r+1}}\right) = \frac{16}{^{16}C_{r+1}} \] Thus, 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 simplification and cancellation, the expression reduces to: \[ \frac{1^{13}}{^{14}C_0\cdot {}^{14}C_1\cdots {}^{14}C_{12}} \] Hence, \[ \alpha = 1 \]
Step 3: Compute the required value. \[ 30\alpha = 30 \times 1 = \boxed{30} \]