1. Key Concepts and Formulas

This problem involves summations of binomial coefficients multiplied by powers of $r$. The key to solving such problems efficiently is to utilize fundamental identities involving binomial coefficients:

• Identity 1: Relationship between consecutive binomial coefficients $r \cdot {}^{n}{C_r} = n \cdot {}^{n - 1}{C_{r - 1}}$ This identity is crucial for simplifying terms like $r \cdot {}^{n}{C_r}$ or $r^2 \cdot {}^{n}{C_r}$ by reducing the upper index of the binomial coefficient. It is derived from the definition ${}^{n}{C_r} = \frac{n!}{r!(n-r)!}$. $r \cdot \frac{n!}{r!(n-r)!} = r \cdot \frac{n \cdot (n-1)!}{r \cdot (r-1)! (n-r)!} = n \cdot \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!} = n \cdot {}^{n-1}{C_{r-1}}$ This identity holds for $r \ge 1$. For $r=0$, both sides are zero.

• Identity 2: Sum of binomial coefficients $\sum_{k=0}^{m} {}^{m}{C_k} = 2^m$ This represents the sum of all binomial coefficients for a given $m$, which is a direct consequence of the binomial theorem $(1+1)^m$.

2. Problem Statement

We are given the equation: $\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}}$ Our goal is to find the value of $\alpha$.

3. Step-by-Step Solution

Let the given summation be $S$. We need to evaluate $S$ and then compare it to the given expression to find $\alpha$.

$S = \sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r}}$

Step 3.1: Apply the first binomial identity to simplify $r^2 \cdot {}^{2023}{C_r}$

The term ${r^2}{}~^{2023}{C_r}$ can be rewritten by applying the identity $r \cdot {}^{n}{C_r} = n \cdot {}^{n - 1}{C_{r - 1}}$. We can write $r^2 \cdot {}^{2023}{C_r} = r \cdot (r \cdot {}^{2023}{C_r})$. Applying the identity with $n=2023$: $r \cdot {}^{2023}{C_r} = 2023 \cdot {}^{2022}{C_{r - 1}}$ Substitute this back: $S = \sum\limits_{r = 0}^{2023} {r \cdot \left( {2023 \cdot {}^{2022}{C_{r - 1}}} \right)}$ $S = 2023 \sum\limits_{r = 0}^{2023} {r \cdot {}^{2022}{C_{r - 1}}}$

• Explanation: We factored out the constant $2023$. Note that for $r=0$, the term $r^2 \cdot {}^{2023}{C_r}$ is $0$. Similarly, $r \cdot {}^{2022}{C_{r-1}}$ is $0$ for $r=0$ (since ${}^{2022}{C_{-1}}$ is taken as 0 by convention), so the summation effectively starts from $r=1$ without loss of generality.

Step 3.2: Decompose the $r$ term and split the summation

Inside the summation, we have $r \cdot {}^{2022}{C_{r - 1}}$. To apply our identity again, we need a factor of $(r-1)$ in front of ${}^{2022}{C_{r - 1}}$. We can achieve this by rewriting $r$ as $(r-1)+1$. $S = 2023 \sum\limits_{r = 0}^{2023} {[(r - 1) + 1]\,.\,{}^{2022}{C_{r - 1}}}$ Now, we split the summation into two parts: $S = 2023 \left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{}^{2022}{C_{r - 1}} + \sum\limits_{r = 0}^{2023} {{}^{2022}{C_{r - 1}}} } } \right]$

• Explanation: This decomposition is a standard technique when you have an $r$ term multiplying a binomial coefficient with $(r-1)$ in its lower index. It allows us to prepare the first sum for another application of Identity 1 and leaves the second sum in a simpler form ready for Identity 2.

Step 3.3: Evaluate the first summation term

Consider the first sum: $\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{}^{2022}{C_{r - 1}}}$.

• For $r=0$, the term is $(-1) \cdot {}^{2022}{C_{-1}} = 0$.
• For $r=1$, the term is $(1-1) \cdot {}^{2022}{C_{0}} = 0 \cdot 1 = 0$. So, this summation effectively starts from $r=2$. Let $k = r-1$. The sum becomes $\sum\limits_{k = 1}^{2022} {k\,.\,{}^{2022}{C_{k}}}$. Now, apply Identity 1 ($k \cdot {}^{m}{C_k} = m \cdot {}^{m-1}{C_{k-1}}$) with $m=2022$: $(r - 1)\,.\,{}^{2022}{C_{r - 1}} = 2022 \cdot {}^{2021}{C_{r - 2}}$ Substituting this back into the first sum (and adjusting the limits from $r=2$): $\sum\limits_{r = 2}^{2023} {2022\,.\,{}^{2021}{C_{r - 2}}} = 2022 \sum\limits_{r = 2}^{2023} {{}^{2021}{C_{r - 2}}}$ Let $j = r-2$. When $r=2, j=0$. When $r=2023, j=2021$. $= 2022 \sum\limits_{j = 0}^{2021} {{}^{2021}{C_{j}}}$ Now, apply Identity 2: $\sum_{j=0}^{2021} {}^{2021}{C_j} = 2^{2021}$. So, the first summation term evaluates to: $2022 \cdot {2^{2021}}$
• Explanation: We again used the key identity to reduce the complexity of the term. Carefully shifting the summation index (from $r$ to $k$ then to $j$) is crucial to correctly identify the form for the sum of binomial coefficients.

Step 3.4: Evaluate the second summation term

Consider the second sum: $\sum\limits_{r = 0}^{2023} {{}^{2022}{C_{r - 1}}}$.

• For $r=0$, the term is ${}^{2022}{C_{-1}} = 0$. So, this summation effectively starts from $r=1$. Let $k = r-1$. When $r=1, k=0$. When $r=2023, k=2022$. $\sum\limits_{k = 0}^{2022} {{}^{2022}{C_{k}}}$ Apply Identity 2: $\sum_{k=0}^{2022} {}^{2022}{C_k} = 2^{2022}$. So, the second summation term evaluates to: ${2^{2022}}$
• Explanation: This is a direct application of the sum of binomial coefficients formula after a simple change of index.

Step 3.5: Combine and Simplify

Now substitute the evaluated summation terms back into the expression for $S$: $S = 2023\left[ {2022 \cdot {2^{2021}} + {2^{2022}}} \right]$ To simplify, we can factor out a common term, $2^{2021}$, from inside the bracket: $S = 2023\left[ {2022 \cdot {2^{2021}} + 2 \cdot {2^{2021}}} \right]$ $S = 2023 \cdot {2^{2021}} \left[ {2022 + 2} \right]$ $S = 2023 \cdot {2^{2021}} \cdot 2024$

• Explanation: Algebraic simplification helps us consolidate terms and prepare the expression for comparison with the target form. Factoring out $2^{2021}$ makes the addition inside the bracket straightforward.

Step 3.6: Determine the value of $\alpha$

We have calculated $S = 2023 \cdot {2^{2021}} \cdot 2024$. The problem states that $S = 2023 \times \alpha \times {2^{2022}}$. Let's equate the two expressions: $2023 \cdot {2^{2021}} \cdot 2024 = 2023 \times \alpha \times {2^{2022}}$ Divide both sides by $2023$: ${2^{2021}} \cdot 2024 = \alpha \times {2^{2022}}$ To solve for $\alpha$, we need to isolate it. We can rewrite ${2^{2022}}$ as $2 \cdot {2^{2021}}$: ${2^{2021}} \cdot 2024 = \alpha \times (2 \cdot {2^{2021}})$ Now, divide both sides by $2^{2021}$: $2024 = \alpha \times 2$ $\alpha = \frac{2024}{2}$ $\alpha = 1012$

• Explanation: We manipulated the equation to match the form given in the problem statement. The goal is to isolate $\alpha$ by dividing common factors from both sides.

4. Tips and Common Mistakes

• Handling Summation Limits: Be extremely careful with the lower limits of summation when applying identities. Terms like ${}^nC_r$ are zero for $r<0$ or $r>n$. Similarly, terms like $r \cdot {}^{n}{C_r}$ are zero for $r=0$. Adjusting the summation range appropriately (e.g., from $r=0$ to $r=1$ or $r=2$) is vital for correctness.
• Choosing the Right Identity: For sums involving $r \cdot {}^nC_r$ or $r^2 \cdot {}^nC_r$, the identity $r \cdot {}^nC_r = n \cdot {}^{n-1}{C_{r-1}}$ is typically the most direct approach. Using generating functions (differentiation of the binomial expansion) is an alternative, more generalized method.
• Algebraic Precision: Keep track of constants and powers of 2. A small error in factorization or division can lead to an incorrect final answer.

5. Summary and Key Takeaway

This problem demonstrates a classic approach to evaluating summations involving products of $r$ (or $r^2$) and binomial coefficients. By systematically applying the identity $r \cdot {}^{n}{C_r} = n \cdot {}^{n - 1}{C_{r - 1}}$ and the sum of binomial coefficients $\sum_{k=0}^{m} {}^{m}{C_k} = 2^m$, the complex summation can be reduced to a simple algebraic expression. Careful manipulation of summation limits and factorization are essential for arriving at the correct answer.

The value of $\alpha$ is $\mathbf{1012}$.