Question

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

• A.
• B.
• C. 0
• D.

Answer 1

The correct option is B

Consider the expansion of $(1+x{)}^{20}$

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

Put x = -1

⇒ ${}^{20}{C}_0$ - ${}^{20}{C}_1$ + ${}^{20}{C}_2$ .............+${}^{20}{C}_9{+}^{20}{C}_{10}{-}^{20}{C}_{11}{+}^{20}{C}_{12}$ + ............ + ${}^{20}{C}_{20}$

= $({-}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2...........{-}^{20}{C}_9){+}^{20}{C}_{10}+({-}^{20}{C}_{11}{+}^{20}{C}_{12}+..............{+}^{20}{C}_{20})$

${}^{20}{C}_0{=}^{20}{C}_{20}{,}^{20}{C}_1{=}^{20}{C}_{19},........{.}^{20}{C}_9{=}^{20}{C}_{11}$

⇒ $20=2\times {(}^{20}{C}_0{-}^{20}{C}_1+........{-}^{20}{C}_9){+}^{20}{C}_{10}$

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

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

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