Animated Solution for Mathematics - Binomial Theorem: If $C_r$ stands for ${}^n{C}_r$, then the sum of the series $\frac{2(\frac{n}{2})!(\frac{n}{2})!}{n!}[C_0^2-2C_1^2+3C_2^2-....+(-1{)}^n(n+1)C_n^2]$, where $n$ is an even positive integer, is equal to

Taming the Binomial Beast

Welcome, fellow traveler on this journey through the elegant world of combinatorics. When you first look at a problem like this, it is natural to feel a sense of intimidation.

You see a series:

It looks like a wall of symbols, a fortress of factorials. But remember, in JEE Advanced mathematics, complexity is often just a mask for a hidden, beautiful simplicity. Our goal today is not to fight this problem with brute force, but to dismantle it piece by piece.

The Power of Decomposition

Let us focus on the heart of the expression: the series inside the square brackets. Let us call this series $S$. We can write it in compact summation notation as:

Now, here is the first secret: never try to solve a complex term like $(r+1)C_r^2$ all at once. It is a product, and products in summations are often traps.

Instead, let us use the distributive property. We can split this single summation into two distinct, manageable parts:

By doing this, we have transformed one 'monster' into two 'pets.' We can train them individually.

The Symmetry Trick

Let us look at the first part, $S_1={\sum }_{r=0}^n(-1{)}^{r}r{C}_r^2$. This looks tricky because of that $r$ sitting there. But we have a powerful tool in our arsenal: the symmetry of binomial coefficients, where $C_r=C_{n-r}$.

Imagine we replace $r$ with $(n-r)$ throughout the summation. The sum remains the same because we are just adding the terms in reverse order. So, $S_1$ can also be written as:

Since $n$ is even, $(-1{)}^{n-r}$ simplifies beautifully to $(-1{)}^{-r}$, which is the same as $(-1{)}^r$. And since $C_{n-r}=C_r$, our expression becomes:

If we expand this, we get $n\sum (-1{)}^r{C}_r^2-\sum (-1{)}^{r}r{C}_r^2$. Notice something? The second part is just our original $S_1$!

We have created a recursive relationship:

This is the 'Aha!' moment. We can solve for $S_1$ algebraically: $2S_1=n\sum (-1{)}^r{C}_r^2$, which means:

We have tamed the first part of our series!

The Grand Finale

Now, we need the value of the second part, ${\sum }_{r=0}^n(-1{)}^r{C}_r^2$. This is a standard identity for even $n$, evaluating to:

Combining everything back into $S$, we get:

Factoring out the common terms, we find:

Finally, we multiply this by the external factor provided in the question: $\frac{2(\frac{n}{2})!(\frac{n}{2})!}{n!}$. When you multiply this by our result for $S$, watch the magic happen.

The factorials in the denominator of the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{\frac{n}{2}}\right)$ are $\frac{n!}{(\frac{n}{2})!(\frac{n}{2})!}$. These perfectly cancel with the external factor. The $2$ in the numerator cancels with the $2$ in the denominator of $(\frac{n}{2}+1)$.

Everything vanishes, leaving us with the elegant, simple answer:

This is the beauty of mathematics. We started with a complex, intimidating expression, and through the systematic application of symmetry and algebraic properties, we stripped away the noise to reveal a clean, perfect result. Never fear the complexity; trust the process, and the solution will reveal itself.