How find the value of $(a_0-a_2+a_4-\ldots)^2+(a_1-a_3+ \ldots)^2$ using $(1+x)^n=a_0+a_1x+a_2x^2+\ldots+a_nx^n$?

Question

Q) $(1+x)^n=a_0+a_1x+a_2x^2+\ldots+a_nx^n$ then $(a_0-a_2+a_4-\ldots)^2+(a_1-a_3+ \ldots)^2$ is equals to
1. 1
2. 0 (zero)
3. $2^{n-1}$
4. $2^n$

Answer: (4)

well this time i am rocked by this question... sorry for that

can you please give me a hint or a certain logic using which i could solve this question?

Answer 1

Hint

\begin{align*} (1+x)^n & = a_0+a_1x+a_2x^2+a_3x^3+ \dotsb +a_nx^n\\ (1-x)^n & = a_0-a_1x+a_2x^2-a_3x^3+ \dotsb +(-1)^na_nx^n\\ \end{align*} Then $$\frac{(1+x)^n+(1-x)^n}{2} = a_0+a_2x^2+a_4x^4+\dotsb$$ Now substitute $x=i$ to get \begin{align*} \frac{(1+i)^n+(1-i)^n}{2} & = a_0-a_2+a_4+\dotsb \\ (\sqrt{2})^n \cos \frac{n\pi}{4} & = a_0-a_2+a_4+\dotsb \\ 2^{n} \cos^2 \frac{n\pi}{4} & = (a_0-a_2+a_4+\dotsb)^2 \\ \end{align*} Now take a similar approach to get the second expression.