Question

The sum of the series ${}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3+.....-......{+}^{20}{C}_{10}$ is-

• A. $\frac{1}{2}{.}^{20}{C}_{10}$
• B. $0$
• C. ${-}^{20}{C}_{10}$
• D. ${}^{20}{C}_{10}$

Answer 1

The correct option is A $\frac{1}{2}{.}^{20}{C}_{10}$

${(1-x)}^{20}{=}^{20}{C}_0{-}^{20}{C}_{1}x{+}^{20}{C}_2{x}^2{-}^{20}{C}_3{x}^3.....{+}^{20}{C}_{10}{-}^{20}{C}_{11}{+}^{20}{C}_{12}......{+}^{20}{C}_{20}$

We know, ${}^n{C}_r{=}^n{C}_{n-r}$

${\therefore }^{20}{C}_{11}{=}^{20}{C}_9$

${}^{20}{C}_{12}{=}^{20}{C}_8$

$\Rightarrow {(1-x)}^{20}{=}^{20}{C}_0{-}^{20}{C}_{1}x.....{+}^{20}{C}_{10}{x}^{10}....{.}^{20}{C}_9{x}^{11}{+}^{20}{C}_8{x}^{12}....{.}^{20}{C}_1{x}^{19}{+}^{20}{C}_0{x}^{20}$

when $x=1$

$\Rightarrow 0=2({}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2...{.}^{20}{C}_9)+{(}^{20}{C}_{10})$

${\Rightarrow }^{20}{C}_0{-}^{20}{C}_1....{.}^{20}{C}_9={\frac{-1}{2}}^{20}{C}_{10}$

Adding ${}^{20}{C}_{10}$ on both sides

${\Rightarrow }^{20}{C}_0{-}^{20}{C}_1....{.}^{20}{C}_9{+}^{20}{C}_{10}={\frac{1}{2}}^{20}{C}_{10}$ is the correct answer.