Question

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

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

Answer 1

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

We know that ${(1+x)}^{20}{=}^{20}{C}_0{+}^{20}{C}_{1}x+...{+}^{20}{C}_{10}{x}^{10}+...{+}^{20}{C}_{20}{x}^{20}$

Putting $x=-1$, we get

$0{=}^{20}{C}_0{-}^{20}{C}_1+...{-}^{20}{C}_9{+}^{20}{C}_{10}{-}^{20}{C}_{11}+...{+}^{20}{C}_{20}$

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

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

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

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