Question

The value of $\sum _{r=0}^{20}{}^{50-r}C_6$ is equal to

• A. ${}^{51}C_7-{}^{30}C_7$
• B. ${}^{51}C_7+{}^{30}C_7$
• C. ${}^{50}C_7-{}^{30}C_7$
• D. ${}^{51}C_6-{}^{30}C_6$

Answer 1

The correct option is A

${}^{51}C_7-{}^{30}C_7$

The explanation for the correct option

The given expression: $\sum _{r=0}^{20}{}^{50-r}C_6$.

$$\begin{gathered} \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{31}C_6+{}^{30}C_6 \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6=\left({}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{31}C_6+{}^{30}C_6\right)+{}^{30}C_7-{}^{30}C_7 \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{31}C_6+\left({}^{30}C_6+{}^{30}C_7\right)-{}^{30}C_7 \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{31}C_6+\left({}^{31}C_7\right)-{}^{30}C_7\left[\because {}^{n}C_r+{}^{n}C_{r-1}={}^{n+1}C_r\right] \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{32}C_6+\left({}^{31}C_6+{}^{31}C_7\right)-{}^{30}C_7 \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{32}C_6+\left({}^{32}C_7\right)-{}^{30}C_7 \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{33}C_6+\left({}^{32}C_6+{}^{32}C_7\right)-{}^{30}C_7 \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{50}C_6+{}^{49}C_6+{}^{48}C_6+.......+{}^{33}C_6+\left({}^{33}C_7\right)-{}^{30}C_7 \\ \Rightarrow \sum _{r=0}^{20}{}^{50-r}C_6={}^{51}C_7-{}^{30}C_7 \end{gathered}$$

Therefore, the value of $\sum _{r=0}^{20}{}^{50-r}C_6$ is equal to ${}^{51}C_7-{}^{30}C_7$.

Hence, the correct option is (A).