Animated Solution for Mathematics - Binomial Theorem: If $({}^{30}{C}_0{)}^2+2({}^{30}{C}_1{)}^2+3({}^{30}{C}_2{)}^2+\cdots +31({}^{30}{C}_{30}{)}^2=\frac{\alpha 60!}{(30!{)}^2}$, then $\alpha$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering the hidden architecture of binomial coefficients.

When you look at a series like $S={\sum }_{r=0}^{30}r({}^{30}{C}_r{)}^2$, it is easy to feel overwhelmed by the sheer number of terms. But remember, in the world of mathematics, complexity is often just a mask for a deeper, simpler truth. Our goal is to peel back that mask.

A Powerful Transformation

The first step in our journey is to recognize that we are dealing with a sum that has a beautiful, symmetric structure. We define our series as $S={\sum }_{r=0}^{30}r({}^{30}{C}_r{)}^2$.

Now, we invoke one of the most powerful tools in our arsenal: the symmetry property of binomial coefficients, which states that ${}^n{C}_r={}^n{C}_{n-r}$.

Imagine you are standing at the beginning of the series, looking at the terms. If we replace $r$ with $30-r$, we are essentially looking at the series from the other end. Because of the symmetry property, the value of the sum remains identical.

So, we write:

Using our symmetry property, we know that ${}^{30}{C}_{30-r}$ is exactly the same as ${}^{30}{C}_r$. Thus, our equation transforms into:

Now, we have two different ways to write the same sum $S$. This is where the magic happens. If we add these two equations together, we get:

Look closely at the term inside the bracket: $r+30-r$. The $r$ and $-r$ cancel out perfectly, leaving us with a constant $30$.

This is the 'Aha!' moment. The variable that was making our series difficult to sum has vanished! We are left with:

Dividing by $2$, we find that $S=15{\sum }_{r=0}^{30}({}^{30}{C}_r{)}^2$.

We are almost there. We now need to evaluate the sum of the squares of the binomial coefficients. This is a standard identity in combinatorics:

For $n=30$, this becomes ${}^{60}{C}_{30}$. Substituting this back into our equation for $S$, we get:

Finally, we expand ${}^{60}{C}_{30}$ using the factorial formula ${}^n{C}_r=\frac{n!}{r!(n-r)!}$:

Comparing this to the form given in the problem, $\frac{\alpha 60!}{(30!{)}^2}$, it is immediately clear that $\alpha =15$.

You see? What started as a daunting summation became a simple exercise in symmetry and identity application. Never fear the complexity of a problem; instead, look for the underlying patterns.

You have the tools, you have the logic, and now, you have the experience. Keep practicing, keep questioning, and keep falling in love with the elegance of mathematics.