Question Details

$If the product (\frac{1}{15_{C 0}} + \frac{1}{15_{C 1}}) (\frac{1}{15_{C 1}} + \frac{1}{15_{C 2}}) \dots (\frac{1}{15_{C 12}} + \frac{1}{15_{C 13}}) = \frac{13}{14_{C 0} 14_{C 1} 14_{C 2} \dots 14_{C 12}}, then 30 α is equal to$

Options

Show Answer

Correct Answer :

Option B

Solution :

The correct answer is 32.

We are given the following equation:

$(\frac{1}{{}_{15}C_{0}} + \frac{1}{{}_{15}C_{1}}) (\frac{1}{{}_{15}C_{1}} + \frac{1}{{}_{15}C_{2}}) \dots (\frac{1}{{}_{15}C_{12}} + \frac{1}{{}_{15}C_{13}}) = \frac{13^{α}}{{}_{14}C_{0} {}_{14}C_{1} {}_{14}C_{2} \dots {}_{14}C_{12}}$

Let us simplify the general term of the product on the left-hand side for $r = 1, 2, \dots, 13$ :

$T_{r} = \frac{1}{{}_{15}C_{r - 1}} + \frac{1}{{}_{15}C_{r}} = \frac{{}_{15}C_{r} + {}_{15}C_{r - 1}}{{}_{15}C_{r - 1} \cdot {}_{15}C_{r}}$

Using Pascal's identity ${}_{n}C_{r} + {}_{n}C_{r - 1} = {}_{n + 1}C_{r}$ , the numerator becomes:
${}_{15}C_{r} + {}_{15}C_{r - 1} = {}_{16}C_{r}$

Also, using the combination property ${}_{n}C_{r} = \frac{n}{r} {}_{n - 1}C_{r - 1}$ , we have:
${}_{16}C_{r} = \frac{16}{r} {}_{15}C_{r - 1}$

Substituting this into the numerator of $T_{r}$ :

$T_{r} = \frac{\frac{16}{r} {}_{15}C_{r - 1}}{{}_{15}C_{r - 1} \cdot {}_{15}C_{r}} = \frac{16}{r \cdot {}_{15}C_{r}}$

Next, using the relation $r \cdot {}_{15}C_{r} = 15 \cdot {}_{14}C_{r - 1}$ , we can rewrite $T_{r}$ as:

$T_{r} = \frac{16}{15 \cdot {}_{14}C_{r - 1}}$

Now, we take the product of $T_{r}$ for $r = 1$ to $r = 13$ :

$\prod_{r = 1}^{13} T_{r} = \prod_{r = 1}^{13} \frac{16}{15 \cdot {}_{14}C_{r - 1}} = \frac{16^{13}}{15^{13} \cdot ({}_{14}C_{0} {}_{14}C_{1} \dots {}_{14}C_{12})}$

Comparing this result with the given right-hand side $\frac{13^{α}}{{}_{14}C_{0} {}_{14}C_{1} \dots {}_{14}C_{12}}$ (where the base simplifies to powers of 2 for $30 α$ ):