Suppose $$\sum_{r=0}^{2023} r^2 \cdot {^{2023}C_r} = 2023 \times \alpha \times 2^{2022}$$, then the value of $$\alpha$$ is

We need to find $$\alpha$$ where $$\sum_{r=0}^{2023} r^2 \binom{2023}{r} = 2023 \times \alpha \times 2^{2022}$$.

We know that $$\sum_{r=0}^{n} r\binom{n}{r} = n \cdot 2^{n-1}$$ and that $$\sum_{r=0}^{n} r^2\binom{n}{r} = n(n+1) \cdot 2^{n-2}$$. The latter follows from the identity $$r^2 = r(r-1) + r$$ and linearity of summation.

Indeed, by writing $$\sum r^2\binom{n}{r} = \sum r(r-1)\binom{n}{r} + \sum r\binom{n}{r}$$ we obtain $$n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} = n(n-1) \cdot 2^{n-2} + 2n \cdot 2^{n-2} = n(n+1) \cdot 2^{n-2}$$.

Substituting $$n = 2023$$ gives $$\sum_{r=0}^{2023} r^2\binom{2023}{r} = 2023 \times 2024 \times 2^{2021}$$.

Comparing this with $$2023 \times \alpha \times 2^{2022}$$ leads to the equation $$2023 \times 2024 \times 2^{2021} = 2023 \times \alpha \times 2^{2022}\,$$ which simplifies to $$2024 \times 2^{2021} = \alpha \times 2^{2022}$$ and hence $$\alpha = \frac{2024}{2} = 1012$$.

The answer is 1012.