"Suppose $$\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}} $$. Then the value of $\alpha$ is ___________"

Question

Accepted Answer

"

Concept :

(1) ${}^n{C_r} = {n \over r}\,.\,{}^{n - 1}{C_{r - 1}}$

Given,

$\sum\limits_{r = 0}^{2023} {{r^2}\,.\,{}^{2023}{C_r}}$

$= \sum\limits_{r = 0}^{2023} {{r^2}\,.\,{{2023} \over r}\,.{}^{2022}{C_{r - 1}}}$

$= 2023\sum\limits_{r = 0}^{2023} {{r}\,.\,{}^{2022}{C_{r - 1}}}$

$= 2023\sum\limits_{r = 0}^{2023} {[(r - 1) + 1]\,.\,{}^{2022}{C_{r - 1}}}$

$$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{}^{2022}{C_{r - 1}} + \sum\limits_{r = 0}^{2023} {{}^{2022}{C_{r - 1}}} } } \right]$$

$$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{{2022} \over {(r - 1)}}\,.\,{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$$

$$ = 2023\left[ {2022\sum\limits_{r = 0}^{2023} {{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$$

$= 2023\left[ {2022\,.\,{2^{2021}} + {2^{2022}}} \right]$

$= 2023\,.\,{2^{2021}}\left[ {2022 + 2} \right]$

$= 2023\,.\,{2^{2021}}\,.\,2024$

$= 2023\,.\,{{{2^{2022}}} \over {{2}}}\,.\,2024$

$= 2023\,.\,{2^{2022}}\,.\,{1012}$

$\therefore$ $\alpha = {1012}$

"

Suppose $$\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}} $$. Then the value of $\alpha$ is ___________