Question

Suppose $\displaystyle\sum_{r=0}^{2023} r^{22023} C_r=2023 \times \alpha \times 2^{2022}$ Then the value of $\alpha$ is

Answer 1

Correct Answer: 1012

The correct answer is 1012.
using result
$\sum _{r=0}^n{r}^2{}^n{C}_r=n(n+1)\cdot 2^{n-2}$
Then $\sum _{r=0}^{2023}{r}^2{}^{2023}{C}_r=2023\times 2024\times 2^{2021}$
$=2023\times \alpha \times 2^{2022}\text{So,}$
$\Rightarrow \alpha =1012$

Answer 2

Using the result: \[ \sum_{r=0}^{n} \binom{n}{r} \cdot C_r = n \cdot (n+1) \cdot 2^{n-2}, \] for \( n = 2023 \), we have: \[ \sum_{r=0}^{2023} \binom{2023}{r} \cdot C_r = 2023 \cdot 2024 \cdot 2^{2021}. \] Equating: \[ 2023 \cdot \alpha \cdot 2^{2022} = 2023 \cdot 2024 \cdot 2^{2021}. \] Simplify: \[ \alpha = \frac{2024}{2} = 1012. \] Final Answer: 1012.