Animated Solution for Mathematics - Binomial Theorem: Let $\alpha ={\sum }_{k=0}^n\frac{{(}^n{C}_k{)}^2}{k+1}$ and $\beta ={\sum }_{k=0}^{n-1}\frac{{}^n{C}_k^n{C}_{k+1}}{k+2}$. If $5\alpha =6\beta$, then $n$ equals

Imagine you are standing before a daunting mountain of algebra. You see two summations, $\alpha$ and $\beta$, filled with binomial coefficients and denominators that seem designed to block your path.

In the world of JEE Advanced, these are not obstacles; they are puzzles waiting for the right key. Today, we are going to unlock them together.

The Absorption Identity

The first thing that should catch your eye is the denominator. In $\alpha$, we have $k+1$ under ${(}^n{C}_k{)}^2$. In $\beta$, we have $k+2$ under the product of two binomial coefficients.

Whenever you see a binomial coefficient divided by its index plus one, you should immediately think of the Absorption Identity:

This identity is our magic wand. It allows us to 'absorb' that pesky denominator into the binomial coefficient itself, transforming a fraction into a cleaner, more manageable form.

For $\alpha$, we split the square term ${(}^n{C}_k{)}^2$ into ${}^n{C}_k{\cdot }^n{C}_k$. We take one ${}^n{C}_k$ and the $k+1$ from the denominator to replace it with $\frac{{}^{n+1}{C}_{k+1}}{n+1}$. Since $n+1$ is a constant, we can pull it right out of the summation.

Now that we have simplified the expressions, we face a product of two binomial coefficients. This is the classic setup for Vandermonde's Identity.

However, Vandermonde's Identity requires the sum of the lower indices to be a constant. In our current form, the lower indices are $k+1$ and $k$, which is not a constant. This is where the Symmetry Property comes to our rescue:

By replacing the second binomial coefficient with its symmetric counterpart, the lower indices become $(k+1)$ and $(n-k)$. When we add them together, the $k$ terms cancel out perfectly, leaving us with $n+1$.

Now, the sum of the lower indices is a constant!

Vandermonde's Identity

With the lower indices aligned, we invoke the power of Vandermonde's Identity:

For $\alpha$, our sum becomes ${\frac{1}{n+1}}^{2n+1}{C}_{n+1}$. For $\beta$, following the same logic, we arrive at ${\frac{1}{n+1}}^{2n+1}{C}_{n+2}$.

The monstrous summations have collapsed into two elegant, single terms.

We are left with the condition $5\alpha =6\beta$. Substituting our simplified expressions, the common factor $\frac{1}{n+1}$ cancels out immediately, leaving us with a ratio of two binomial coefficients:

Expanding these into factorials is the final step. As you write out the factorials, watch how the $(2n+1)!$ terms vanish, and the remaining factorials simplify beautifully.

You will find yourself with the simple linear equation:

Solving this gives us $n=10$. You see? What looked like an impossible mountain was just a series of elegant steps. Keep this logic in your toolkit, and no binomial problem will ever intimidate you again.