Question

${(}^{10}{C}_0{)}^2-{(}^{10}{C}_1{)}^2+....-{(}^{10}{C}_9{)}^2+{(}^{10}{C}_{10}{)}^{2}equals$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$(1+x{)}^{10}{=}^{10}{C}_0{+}^{10}{C}_1{+}^{10}{C}_2{x}^2+....{+}^{10}{C}_{10}{x}^{10}...(i)Also,(1-x{)}^{10}{=}^{10}{C}_0{-}^{10}{C}_{1}x{+}^{10}{C}_2{x}^2+...{.}^{10}{C}_{10}{x}^{10}Multiplying,weget(1-x^2{)}^{10}={(}^{10}{C}_0{+}^{10}{C}_{1}x{+}^{10}{C}_2{x}^2+....{+}^{10}{C}_{10}{x}^{10})\times {(}^{10}{C}_0{-}^{10}{C}_{1}x{+}^{10}{C}_2{x}^2+....{+}^{10}{C}_{10}{x}^{10})Equatingthecoefficientsof{x}^{10},weget{}^{}{}^{10}{C}_5(-1{)}^5{=}^{10}{C}_0{}^{10}{C}_{10}{-}^{10}{C}_1{}^{10}{C}_9{+}^{10}{C}_2{}^{10}{C}_8+....{+}^{10}{C}_{10}{}^{10}{C}_0\Rightarrow {-}^{10}{C}_5={(}^{10}{C}_0{)}^2-{(}^{10}{C}_1{)}^2+{(}^{10}{C}_2{)}^2-....{(}^{10}{C}_{10}{)}^2$