Question

The sum of the series ${}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3+-..{+}^{20}{C}_{10}$ is${\frac{1}{2}}^{20}{C}_{10}$

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

Answer 1

Answer: (D)

${\frac{1}{2}}^{20}{C}_{10}$

The sum of the series

$\begin{array}{c}{}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3+....{.}^{20}{C}_{10}\\ {\left(1+x\right)}^{20}{=}^{20}{C}_0{+}^{20}{C}_{1}x{+}^{20}{C}_2{x}^2+...{.}^{20}{C}_{10}{x}^{10}\\ putx=-1\\ {\left(1-1\right)}^{20}{=}^{20}{C}_0{+}^{20}{C}_1{+}^{20}{C}_2+......{+}^{20}{C}_{10}\\ 0{=}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2+.......{.}^{20}{C}_{10}\\ =2\left[{}^{20}{C}_0{-}^{20}{C}_1+{.....}^{20}{C}_9\right]{+}^{20}{C}_{10}\\ ={\frac{1}{2}}^{20}{C}_{10}\end{array}$

Hence, this is the answer.