Question

We know that if ${}^n{C}_0,{}^n{C}_1,{}^n{C}_2,\dots ,{}^n{C}_n$ be binomial coefficients then $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\dots +C_n{x}^n$. Various relations among binomial coefficients can be derived by putting $x=1,-1,x=i,x=w,\text{where,}i=\sqrt{-1},w=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$. Some other identities can be derived by adding and subtracting two such identities. The expression $(a+ib{)}^n$ can be evaluated by using De-Moiver's theorem by putting
$a=r\cos \theta ,b=r\sin \theta$.

The value of the expression ${(}^n{C}_0-{}^n{C}_2+{}^n{C}_4-{}^n{C}_6+\dots {)}^2+{(}^n{C}_1-{}^n{C}_3+{}^n{C}_5\dots {)}^2$ must be

• A. $2^{2n}$
• B. $2^n$
  
• C. $2^{n^2}$
  
• D. $2^{n/2}$

Answer 1

The correct option is B $2^n$


$(1+i{)}^n=C_0+C_{1}i+C_2{i}^2+C_3{i}^3+...+C_n{i}^n$
$=(C_0-C_2+C_4-C_6+...+i(C_1-C_3+C_5-...$

Taking modulus, we get,
$(C_0-C_2+C_4-C6+...{)}^2+(C_1-C_3+C_5–...{)}^2={\left({\left(\sqrt{2}\right)}^n\right)}^2=2^n$