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 $\frac{1}{2}{20}_{C_{10}}$

We know that,

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

Putting $x=-1$ we get

$0{=}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3{+}_{......}{}_{......}{+}^{20}{C}_{10}{+}_{.............}{+}^{20}{C}_{20}$

But as we know , that ${}^{20}{C}_0{=}^{20}{C}_{20}{,}^{20}{C}_1{=}^{20}{C}_{19}$ and so on , we can write above equation as

$0=2{(}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3+..........{-}^{20}{C}_9){+}^{20}{C}_{10}-\frac{{}^{20}{C}_{10}}{2}={(}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3+..........{-}^{20}{C}_9)$

Hence,

${}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3{+}_{......}{-}_{......}{+}^{20}{C}_{10}=-\frac{{}^{20}{C}_{10}}{2}{+}^{20}{C}_{10}=\frac{{}^{20}{C}_{10}}{2}$