Animated Solution for Mathematics - Binomial Theorem: Suppose ${\sum }_{r=0}^{2023}{r}^2\cdot {}^{2023}{C}_r=2023\times \alpha \times 2^{2022}$. Then the value of $\alpha$ is ______.

When you first look at a problem like ${\sum }_{r=0}^{2023}{r}^2\cdot {}^{2023}{C}_r=2023\times \alpha \times 2^{2022}$, it is natural to feel a shiver of hesitation. The number $2023$ is large, and the summation looks like a mountain of arithmetic.

But in the world of JEE Advanced, we do not climb mountains by brute force; we climb them by finding the hidden paths. This problem is not about calculation; it is about recognizing the elegant structure of binomial identities.

The core of this problem lies in the standard identity for the sum of $r^2$ multiplied by a binomial coefficient. Specifically, we are looking at ${\sum }_{r=0}^n{r}^2\cdot {}^n{C}_r$.

This is a classic result that every aspirant should have in their mental toolkit. The formula is:

Why does this work? It comes from the binomial expansion $(1+x{)}^n={\sum }_{r=0}^n{}^n{C}_r{x}^r$.

If you differentiate this once, you get a term with $r$. If you multiply by $x$ and differentiate again, you get a term with $r^2$. Setting $x=1$ at the end gives us this beautiful, compact result.

Now, let us apply this to our specific problem. By comparing our given summation ${\sum }_{r=0}^{2023}{r}^2\cdot {}^{2023}{C}_r$ with the general identity, it becomes crystal clear that $n=2023$.

We do not need to expand anything. We simply plug $n=2023$ into our formula:

This simplifies to $2023\times 2024\times 2^{2021}$. This is the moment where many students panic because the powers of $2$ do not match the right-hand side of the original equation. But stay calm; this is where the artistry of math comes in.

We have $2023\times 2024\times 2^{2021}$, and we need to match it with $2023\times \alpha \times 2^{2022}$. Notice that $2024$ is an even number. We can write $2024$ as $1012\times 2$.

Now, substitute this back into our expression:

Using the laws of exponents, $2^1\times 2^{2021}$ becomes $2^{2022}$. So, our expression is now $2023\times 1012\times 2^{2022}$.

Comparing this directly with $2023\times \alpha \times 2^{2022}$, the value of $\alpha$ is staring us in the face. It is $1012$.

The mountain has been climbed, not by brute force, but by the elegance of the binomial theorem. Keep practicing these patterns, and you will find that even the most intimidating problems become simple puzzles waiting to be solved.