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 $k=$

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Aicharya · Accepted Answer

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ight) & +\ldots \ & +i\left(a_1-a_3+a_5-a_7+\ldotsight) \ \Rightarrow\left[\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}ight)ight]^n= & \left(a_0-a_2+a_4-a_6+\ldotsight) \ & +i\left(a_1-a_3+a_5-a_7+\ldotsight) \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

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 $k=$

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