If Σ(k=1 to 31) (31Ck)(31Ck-1) - Σ(k=1 to 30) (30Ck)(30Ck-1) = α(60!)/((30!)(31!)), Where α ∈ R, the

Q: Wait, why does $\sum_{k=1}^{n} \binom{n}{k}\binom{n}{k-1}$ equal $\binom{2n}{n-1}$? I don't see the connection.

Think of it as selecting a committee of $n+1$ people from a group of $2n$ people (split into two groups of $n$). To choose $n+1$ people, you can choose $k$ from the first group and $n+1-k$ from the second. Since $\binom{n}{n+1-k} = \binom{n}{k-1}$, the sum $\sum \binom{n}{k}\binom{n}{k-1}$ counts the total ways to choose $n+1$ people from $2n$, which is simply $\binom{2n}{n+1}$ or $\binom{2n}{2n-(n+1)} = \binom{2n}{n-1}$. It is a beautiful combinatorial identity!

Q: Can I just expand the summations manually? There are only 31 terms.

Technically, yes, but that is a trap! In JEE Advanced, time is your most precious resource. Manual expansion is prone to calculation errors and will consume valuable minutes. Always look for the underlying identity first. The goal is to solve the problem with elegance, not brute force.

Q: Why did we convert everything to factorials at the end? Is there a faster way?

Converting to factorials is the 'universal language' of binomial coefficients. When you have an equation involving $\alpha$ and factorials on the right-hand side, matching the structure of the left-hand side to the right-hand side is the most reliable path. It ensures you don't miss any factors during simplification.

Question

If $\sum_{k=1}^{31} ({}^{31}C_k)({}^{31}C_{k-1}) - \sum_{k=1}^{30} ({}^{30}C_k)({}^{30}C_{k-1}) = \frac{\alpha (60!)}{(30!)(31!)}$, where $\alpha \in \mathbb{R}$, then the value of $16\alpha$ is equal to

Options:

(A) 1411

(B) 1320

(C) 1615

(D) 1855

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion.

When you first look at the expression

$$\sum_{k=1}^{31} \binom{31}{k}\binom{31}{k-1} - \sum_{k=1}^{30} \binom{30}{k}\binom{30}{k-1}$$

it is natural to feel a surge of anxiety. It looks like a mountain of arithmetic, but in the world of JEE Advanced, intimidation is often a mask for a hidden, elegant simplicity.

### The Hidden Identity

The core of this problem lies in recognizing a pattern. We are dealing with the sum of products of consecutive binomial coefficients of the form

$$\sum_{k=1}^{n} \binom{n}{k}\binom{n}{k-1}$$

This is a fundamental identity in combinatorics. By Vandermonde's Identity, this sum is equivalent to

$$\binom{2n}{n-1}$$

By applying this identity, we instantly collapse those complex summations into single terms:

$$\binom{62}{30} - \binom{60}{29}$$

### The Factorial Dance

Now that we have reduced the problem to

$$\binom{62}{30} - \binom{60}{29} = \frac{\alpha (60!)}{(30!)(31!)}$$

we enter the second phase: the algebraic dance. We must translate our combinations into the language of factorials.

We expand the terms as follows:

$$\binom{62}{30} = \frac{62!}{30!32!} \quad 	ext{and} \quad \binom{60}{29} = \frac{60!}{29!31!}$$

We want to factor out

$$\frac{60!}{30!31!}$$

from both terms. For the first term, we rewrite $62!$ as $62 \cdot 61 \cdot 60!$ and $32!$ as $32 \cdot 31!$, yielding:

$$\frac{60!}{30!31!} \cdot \frac{62 \cdot 61}{32}$$

For the second term, we multiply the numerator and denominator by $30$ to obtain:

$$\frac{60!}{30!31!} \cdot 30$$

### The Final Victory

With the common term factored out, the equation simplifies beautifully:

$$\frac{60!}{30!31!} \left( \frac{62 \cdot 61}{32} - 30 ight) = \frac{\alpha (60!)}{(30!)(31!)}$$

The factorial fraction cancels out entirely, leaving us with a simple linear equation for $\alpha$:

$$\alpha = \frac{62 \cdot 61}{32} - 30$$

Calculating this, we find:

$$\alpha = \frac{3782}{32} - 30 = \frac{3782 - 960}{32} = \frac{2822}{32}$$

Finally, the question asks for $16\alpha$. Multiplying our result by $16$ gives us:

$$\frac{2822}{2} = \mathbf{1411}$$

We have conquered the mountain, not by brute force, but by understanding the underlying structure of the math. Keep this perspective, and no problem will ever be too intimidating again.

Animated Solution for Mathematics - Binomial Theorem: If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)}$ , where $α \in R$ , then the value of $16 α$ is equal to

Select Answer:

Visualized Solution

Analyzing the Given Expression

The Core Binomial Identity

Evaluating the First Summation

Simplifying the First Term

Evaluating the Second Summation

Simplifying the Second Term

Setting Up the Simplified Equation

Expanding the Combinations

Identifying Common Terms

Adjusting the First Term

Adjusting the Second Term

Isolating $α$

Computing the Value of $α$

Finding $16 α$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Hidden Identity

The Factorial Dance

The Final Victory

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