Q50 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics
If $(1+x)^n=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$ and $a_0-a_2+a_4-a_6+\ldots=k \cos \frac{n \pi}{4}$, then $k=$
- A$2^n$
- B$2^{2 n}$
- C$\frac{2^n}{2}$
- Correct answerD$2^{\frac{n}{2}}$
Solution
If $(1+x)^n=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots+a_n x^n$
Put $x=i$, we get
$$
\begin{aligned}
(1+i)^n=\left(a_0-a_2+a_4-a_6+\ldots\right) & +\ldots \\
& +i\left(a_1-a_3+a_5-a_7+\ldots\right) \\
\Rightarrow\left[\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right]^n= & \left(a_0-a_2+a_4-a_6+\ldots\right) \\
& +i\left(a_1-a_3+a_5-a_7+\ldots\right)
\end{aligned}
$$
On comparing the real and imaginary parts, we get
$$
2^{n / 2} \cos \frac{n \pi}{4}=a_0-a_2+a_4-a_6+\ldots
$$
and $2^{n / 2} \sin \frac{n \pi}{4}=a_1-a_3+a_5-a_7+\ldots$
$$
\because \quad a_0-a_2+a_4-a_6+\ldots=k \cos \frac{n \pi}{4} \Rightarrow k=2^{\frac{n}{2}}
$$
Hence, option (d) is correct.
Hence, the correct answer is option: 'D'.