If (1/15C0 + 1/15C1)(1/15C1 + 1/15C2) ... (1/15C12 + 1/15C13) = α¹³/ (14C0 14C1 ... 14C12), then 30α

Q: Wait, why can't I just calculate the values of $^{15}C_0, ^{15}C_1$, etc. and multiply them?

That is a classic trap! In JEE Advanced, you are rarely expected to perform massive arithmetic calculations. The binomial coefficients for $n=15$ are quite large, and multiplying them would be a nightmare. The problem is designed to test your ability to recognize patterns and use algebraic identities to simplify the expression before calculating anything.

Q: How do I derive the identity $\frac{1}{^{n}C_{r-1}} + \frac{1}{^{n}C_r} = \frac{n+1}{n \cdot ^{n-1}C_{r-1}}$?

Great question! You can derive it by expanding the binomial coefficients using the formula $^{n}C_r = \frac{n!}{r!(n-r)!}$. If you write out the reciprocals $\frac{(r-1)!(n-r+1)!}{n!} + \frac{r!(n-r)!}{n!}$ and find a common denominator, you will see that it simplifies beautifully to $\frac{n+1}{n \cdot ^{n-1}C_{r-1}}$. It is a fantastic exercise to try this derivation yourself!

Q: Why does the upper index of the binomial coefficient change from 15 to 14?

That is the magic of the identity! When you combine the reciprocals of two consecutive binomial coefficients with upper index $n$, the resulting expression involves a binomial coefficient with upper index $n-1$. This shift is exactly what allows the product to simplify into the form given on the right-hand side of the equation.

Question

If $(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1})(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}) \cdots (\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}) = \frac{\alpha^{13}}{^{14}C_0 \cdot ^{14}C_1 \cdots ^{14}C_{12}}$, then $30\alpha$ is equal to .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a massive, intimidating mountain. The problem before us is a product of thirteen brackets, each containing the sum of reciprocals of consecutive binomial coefficients:

$$\left( \frac{1}{^{15}C_0} + \frac{1}{^{15}C_1} ight) \left( \frac{1}{^{15}C_1} + \frac{1}{^{15}C_2} ight) \cdots \left( \frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}} ight)$$

At first glance, it looks like a calculation that would take hours. But in the world of JEE, we don't climb the mountain by brute force; we find the hidden path.

### The General Term: Simplifying the Chaos

To conquer this, we must first simplify the chaos. Let's look at a single bracket. We can define a general term $T_r$ for any bracket in this sequence as:

$$T_r = \frac{1}{^{15}C_{r-1}} + \frac{1}{^{15}C_r}$$

Our entire expression is simply the product of these $T_r$ terms as $r$ ranges from $1$ to $13$. By focusing on $T_r$, we transform a massive, overwhelming product into a manageable, single-term analysis.

### The Secret Weapon: The Reciprocal Identity

Now, we need a tool to collapse this sum. There is a beautiful, standard identity in combinatorics:

$$\frac{1}{^{n}C_{r-1}} + \frac{1}{^{n}C_r} = \frac{n+1}{n \cdot ^{n-1}C_{r-1}}$$

This identity is the key to the kingdom. It allows us to take the sum of two reciprocals and turn it into a single, elegant term.

Let's apply this to our problem where $n=15$. Substituting $n=15$ into our identity, we get:

$$T_r = \frac{15+1}{15 \cdot ^{15-1}C_{r-1}} = \frac{16}{15 \cdot ^{14}C_{r-1}}$$

Notice the elegance here: the upper index of the binomial coefficient has shifted from $15$ to $14$. This is exactly what we need to match the structure of the right-hand side of our given equation.

### The Collapse: Constructing the Product

With our simplified $T_r$ in hand, we can now reconstruct the entire product $P$:

$$P = \prod_{r=1}^{13} T_r = \prod_{r=1}^{13} \frac{16}{15 \cdot ^{14}C_{r-1}}$$

Since $\frac{16}{15}$ is a constant, it appears in every one of the $13$ terms. We can pull this constant out of the product:

$$P = \left( \frac{16}{15} ight)^{13} \cdot \prod_{r=1}^{13} \frac{1}{^{14}C_{r-1}}$$

This leaves us with the expression:

$$P = \frac{\left( \frac{16}{15} ight)^{13}}{^{14}C_0 \cdot ^{14}C_1 \cdots ^{14}C_{12}}$$

### The Final Victory

Now, compare this result with the right-hand side given in the problem:

$$\frac{\alpha^{13}}{^{14}C_0 \cdot ^{14}C_1 \cdots ^{14}C_{12}}$$

The denominators are identical! This means the numerators must be equal:

$$\alpha^{13} = \left( \frac{16}{15} ight)^{13}$$

Since the power is odd, we can directly conclude that $\alpha = \frac{16}{15}$. Finally, the question asks for $30\alpha$. Substituting our value:

$$30 	imes \frac{16}{15} = 2 	imes 16 = \mathbf{32}$$

We have reached the summit! The mountain wasn't so high after all; it was just a matter of finding the right path.

Enter Numerical Value:

Visualized Solution

Analyze the Expression Structure

Define the General Term $T_{r}$

The Reciprocal Identity

Substitute $n = 15$ into the Identity

Simplify the General Term

Construct the Full Product $P$

Expand the Product Notation

Compare with Given RHS

Find the Value of $α$

Calculate $30 α$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The General Term

The Secret Weapon

The Collapse

The Final Victory

Similar Questions

If $(^{30} C_{0})^{2} + 2 (^{30} C_{1})^{2} + 3 (^{30} C_{2})^{2} + \dots + 31 (^{30} C_{30})^{2} = \frac{α 60 !}{( 30 ! ) ^{2}}$ , then $α$ is equal to

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Suppose $\sum_{r = 0}^{2023} r^{2} \cdot^{2023} C_{r} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ______.

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The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

Enter Numerical Value:

Visualized Solution

Analyze the Expression Structure

Define the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​

The Reciprocal Identity

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=15 into the Identity

Simplify the General Term

Construct the Full Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Expand the Product Notation

Compare with Given RHS

Find the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 30α