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 the value of $$\alpha$$ given that $$\sum_{r=0}^{2023} r^2 \cdot \binom{2023}{r} = 2023 \times \alpha \times 2^{2022}$$.

We use the identity:

$$\sum_{r=0}^{n} r^2 \binom{n}{r} = n(n+1) \cdot 2^{n-2}$$

Start with the binomial theorem $$(1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r$$.

Differentiate with respect to $$x$$: $$n(1+x)^{n-1} = \sum_{r=1}^n r\binom{n}{r}x^{r-1}$$.

Multiply by $$x$$: $$nx(1+x)^{n-1} = \sum_{r=1}^n r\binom{n}{r}x^r$$.

Differentiate again: $$n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2} = \sum_{r=1}^n r^2\binom{n}{r}x^{r-1}$$.

Set $$x = 1$$: $$n \cdot 2^{n-1} + n(n-1) \cdot 2^{n-2} = \sum_{r=0}^n r^2\binom{n}{r}$$.

Factor: $$n \cdot 2^{n-2}(2 + n - 1) = n(n+1) \cdot 2^{n-2}$$.

$$\sum_{r=0}^{2023} r^2 \binom{2023}{r} = 2023 \times 2024 \times 2^{2021}$$

We need to express this as $$2023 \times \alpha \times 2^{2022}$$. So:

$$2023 \times 2024 \times 2^{2021} = 2023 \times \alpha \times 2^{2022}$$

Dividing both sides by $$2023 \times 2^{2021}$$:

$$2024 = \alpha \times 2$$ $$\alpha = \frac{2024}{2} = 1012$$

The value of $$\alpha$$ is 1012.