JEE Main 2023 — Binomial Theorem Question with Solution

Concept :

(1) $${}^n{C}_r=\frac{n}{r}.{}^{n-1}{C}_{r-1}$$

Given,

$$\sum _{r=0}^{2023}{r}^2.{}^{2023}{C}_r$$

$$=\sum _{r=0}^{2023}{r}^2.\frac{2023}{r}.{}^{2022}{C}_{r-1}$$

$$=2023\sum _{r=0}^{2023}r.{}^{2022}{C}_{r-1}$$

$$=2023\sum _{r=0}^{2023}[(r-1)+1].{}^{2022}{C}_{r-1}$$

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

$$=2023\left[\sum _{r=0}^{2023}(r-1).\frac{2022}{(r-1)}.{}^{2021}{C}_{r-2}+2^{2022}\right]$$

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

$$=2023\left[2022.2^{2021}+2^{2022}\right]$$

$$=2023.2^{2021}\left[2022+2\right]$$

$$=2023.2^{2021}.2024$$

$$=2023.\frac{2^{2022}}{2}.2024$$

$$=2023.2^{2022}.1012$$

$$\therefore$$ $$\alpha =1012$$