Question

If the product \[ \left( \frac{1}{\binom{15}{0}} + \frac{1}{\binom{15}{1}} \right) \left( \frac{1}{\binom{15}{1}} + \frac{1}{\binom{15}{2}} \right) \cdots \left( \frac{1}{\binom{15}{12}} + \frac{1}{\binom{15}{13}} \right) \] is equal to \[ \frac{\alpha^{13}}{\binom{14}{0} \binom{14}{1} \binom{14}{2} \cdots \binom{14}{12}}, \] then \( 30\alpha \) is equal to:

• A. 16
• B. 32
• C. 15
• D. 28

Hint:

When dealing with products and sums of binomial coefficients, look for patterns and symmetries in the terms to simplify the expression. Recognizing standard binomial identities can be very helpful in solving such problems.

Answer 1

The Correct Option is B

Step 1: Express the product in terms of binomial coefficients.
We are given the product as: \[ P = \left( \frac{1}{\binom{15}{0}} + \frac{1}{\binom{15}{1}} \right) \left( \frac{1}{\binom{15}{1}} + \frac{1}{\binom{15}{2}} \right) \cdots \left( \frac{1}{\binom{15}{12}} + \frac{1}{\binom{15}{13}} \right) \] We know the general identity for binomial coefficients: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Each term involves the sum of two reciprocals of binomial coefficients. To simplify the product, we observe that the given product can be interpreted using properties of binomial coefficients and their symmetries.
Step 2: Use the identity for binomial sums.
By applying certain binomial identities and properties of coefficients, the expression for the product simplifies to the form: \[ \frac{\alpha^{13}}{\binom{14}{0} \binom{14}{1} \cdots \binom{14}{12}} \] This leads to the expression for \( \alpha \) being related to a known constant involving factorials.
Step 3: Solve for \( \alpha \).
After simplifying and comparing with the given right-hand side expression, we find that: \[ 30 \alpha = 32 \] Thus, \( \alpha = \frac{32}{30} \), and therefore \( 30 \alpha = 32 \).