Question

${\left({}^{2n}{C}_0\right)}^2-{\left({}^{2n}{C}_1\right)}^2+{\left({}^{2n}{C}_2\right)}^2-\dots +{\left({}^{2n}{C}_{2n}\right)}^2=$

• A. ${}^{2n}{C}_n$
• B. $(-1{)}^n{.}^{2n}{C}_n$
• C. $(-1{)}^{n/2}{.}^n{C}_{n/2}$
• D. $(-1{)}^{n/2}{.}^{2n}{C}_n$

Answer 1

The correct option is B $(-1{)}^n{.}^{2n}{C}_n$

${(1+x)}^{2n}{(1-\frac{1}{x})}^{2n}$

$=[{}^{2n}{C}_0+\left({}^{2n}{C}_1\right)x+\left({}^{2n}{C}_2\right)x^2+...+\left({}^{2n}{C}_{2n}\right)x^{2n}]\times [{}^{2n}{C}_0-\left({}^{2n}{C}_1\right)\frac{1}{x}+\left({}^{2n}{C}_2\right)\frac{1}{x^2}+...+\left({}^{2n}{C}_{2n}\right)\frac{1}{x^{2n}}]$

Independent term of $x$ on RHS

$={\left({}^{2n}{C}_0\right)}^2-{\left({}^{2n}{C}_1\right)}^2+{\left({}^{2n}{C}_2\right)}^2-...+{\left({}^{2n}{C}_n\right)}^2$

LHS$={(1+x)}^{2n}{\left(\frac{x-1}{x}\right)}^{2n}$

$=\frac{1}{x^{2n}}{(1-x^2)}^{2n}$

Independent term of $x$ on the LHS$={(-1)}^n{}^{2n}{C}_n$