Suppose sum limits_r = 0^2023 r^2 ^2023C_r = 2023 x alpha x 2^2022 . Then the value of alpha is ___________

Concept : (1) ^nC_r = n / r . ^n - 1C_r - 1 Given, sum limits_r = 0^2023 r^2 . ^2023C_r = sum limits_r = 0^2023 r^2 . 2023 / r .^2022C_r - 1 = 2023sum limits_r = 0^2023 r . ^2022C_r - 1 = 2023sum limits_r = 0^2023 [(r - 1) + 1] . ^2022C_r - 1 = 2023[ sum limits_r = 0^2023 (r - 1) . ^2022C_r - 1 + sum limits_r = 0^2023 ^2022C_r - 1 ] = 2023[ sum limits_r = 0^2023 (r - 1) . 2022 / (r - 1) . ^2021C_r - 2 + 2^2022 ] = 2023[ 2022sum limits_r = 0^2023 ^2021C_r - 2 + 2^2022 ] = 2023[ 2022 . 2^2021 + 2^2022 ] = 2023 . 2^2021[ 2022 + 2 ] = 2023 . 2^2021 . 2024 = 2023 . 2^2022 / 2 . 2024 = 2023 . 2^2022 . 1012 Therefore alpha = 1012

Suppose sumr 02023 r22023Cr 2023 x alpha x 22022 Then t... | null (Mathematics)

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}

Explanation

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