Question

The value of ${}^{40}{C}_0{+}^{40}{C}_1{+}^{40}{C}_2+...{+}^{40}{C}_{20}$ is

• A. $2^{40}+\frac{40!}{{(20!)}^2}$
• B. $2^{39}-\frac{1}{2}\times \frac{40!}{{(20!)}^2}$
• C. $2^{39}{+}^{40}{C}_{20}$
• D. None of these

Answer 1

Answer: (D)

${}^{40}{C}_0{+}^{40}{C}_1{+}^{40}{C}_2+...{+}^{40}{C}_{20}$ $=\frac{1}{2}[{2.}^{40}{C}_0+{2.}^{40}{C}_1+{2.}^{40}{C}_2+...+{2.}^{40}{C}_{20}]$ $=\frac{1}{2}[{(}^{40}{C}_0{+}^{40}{C}_{40})+{(}^{40}{C}_1{+}^{40}{C}_{39})+...+$ ${(}^{40}{C}_{19}{+}^{40}{C}_{21})+2^{40}{C}_{20}]$ $=\frac{1}{2}[{\{}^{40}{C}_0{+}^{40}{C}_1{+}^{40}{C}_2+...{+}^{40}{C}_{19}{+}^{40}{C}_{20}]$ ${+}^{40}{C}_{21}\}{+}^{40}{C}_{20}]$ $=\frac{1}{2}\left[2^{40}+\frac{40!}{{(20!)}^2}\right]=2^{39}+\frac{1}{2}\frac{40!}{{(20!)}^2}$