Question

The sum ${}^{20}{C}_0{+}^{20}{C}_1{+}^{20}{C}_2+\dots \dots {.}^{20}{C}_{10}$ is equal to

• A. $2^{19}+\frac{20!}{2(10!{)}^2}$
• B. $2^{19}-\frac{1}{2}\cdot \frac{20!}{(10!{)}^2}$
• C. $2^{19}{+}^{20}{C}_{10}$
• D. none of these

Answer 1

Answer: (A)

$2^{19}+\frac{20!}{2(10!{)}^2}$

As we know that $(1+x{)}^{20}{=}^{20}{C}_0{+}^{20}{C}_{1}x{+}^{20}{C}_2{x}^2+\cdots {+}^{20}{C}_{20}{x}^{20}$

Put $x=1$

$⟹2^{20}{=}^{20}{C}_0{+}^{20}{C}_1{+}^{20}{C}_2+\cdots {+}^{20}{C}_{20}$

$⟹2^{20}{=}^{20}{C}_0{+}^{20}{C}_1+\cdots {+}^{20}{C}_{10}{+}^{20}{C}_9+\cdots {+}^{20}{C}_1{+}^{20}{C}_0$

$2^{20}=2{(}^{20}{C}_0{+}^{20}{C}_1{+}^{20}{C}_2+\cdots {+}^{20}{C}_{10}){-}^{20}{C}_{10}$

$⟹{}^{20}{C}_0{+}^{20}{C}_1{+}^{20}{C}_2+\cdots {+}^{20}{C}_{10}=2^{19}+{\frac{1}{2}}^{20}{C}_{10}=2^{19}+\frac{1}{2}\cdot \frac{20!}{(10!{)}^2}$