Suppose _r=0^2023r^2,^2023C_r = 20232^2022.

Question

Suppose $\displaystyle\sum_{r=0}^{2023}r^2\,{}^{2023}C_r = 2023	imes\alpha	imes2^{2022}$. Then the value of $\alpha$ is

Answer: 1012

Accepted Answer

Using the identity $\displaystyle\sum_{r=0}^{n}r^2\binom{n}{r}=n(n+1)2^{n-2}$ with $n=2023$:

$$\sum_{r=0}^{2023}r^2\binom{2023}{r}=2023\times2024\times2^{2021}=2023\times\alpha\times2^{2022}$$

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

Solutions · 1