JEE Main 2023 — Binomial Theorem Question with Solution

Given,

$\sum _{r=0}^6{}^{51-k}C_3$

$={}^{51}C_3+{}^{50}C_3+{}^{49}C_3+{}^{48}C_3+{}^{47}C_3+{}^{46}C_3+{}^{45}C_3$

Now we know that ${}^{n}C_r+{}^{n}C_{r+1}={}^{n+1}C_{r+1}$ or ${}^{n}C_r={}^{n+1}C_{r+1}-{}^{n}C_{r+1}$

Now using the above formula in given expression we get,

$\sum _{r=0}^6{}^{51-k}C_3={}^{51}C_3+{}^{50}C_3+{}^{49}C_3+{}^{48}C_3+{}^{47}C_3+\underset{}{\underbrace{{}^{46}C_3+{}^{46}C_4}}-{}^{45}C_4$

$\Rightarrow \sum _{r=0}^6{}^{51-k}C_3={}^{51}C_3+{}^{50}C_3+{}^{49}C_3+{}^{48}C_3+\underset{}{\underbrace{{}^{47}C_3+{}^{47}C_4}}-{}^{45}C_4$

$\Rightarrow \sum _{r=0}^6{}^{51-k}C_3={}^{51}C_3+{}^{50}C_3+{}^{49}C_3+{}^{48}C_3+{}^{48}C_4-{}^{45}C_4$

$\Rightarrow \sum _{r=0}^6{}^{51-k}C_3={}^{51}C_3+{}^{50}C_3+{}^{49}C_3+{}^{49}C_4-{}^{45}C_4$

$\Rightarrow \sum _{r=0}^6{}^{51-k}C_3={}^{51}C_3+{}^{50}C_3+{}^{50}C_4-{}^{45}C_4$

$\Rightarrow \sum _{r=0}^6{}^{51-k}C_3={}^{51}C_3+{}^{51}C_4-{}^{45}C_4$

$\Rightarrow \sum _{r=0}^6{}^{51-k}C_3={}^{52}C_4-{}^{45}C_4$.