If (1+x)^n = a_0 + a_1 x + a_2 x^2 + ldots + | vedclass

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(D) Given $(1+x)^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_n x^n$. Substitute $x = i$ in the expansion: $(1+i)^n = (a_0 - a_2 + a_4 - a_6 + \ld

Vedclass · Accepted Answer

$2^{\frac{n}{2}}$

Vedclass · Answer

(D) Given $(1+x)^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_n x^n$. Substitute $x = i$ in the expansion: $(1+i)^n = (a_0 - a_2 + a_4 - a_6 + \ldots) + i(a_1 - a_3 + a_5 - a_7 + \ldots)$. Express $1+i$ in polar form: $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$. Using De Moivre's Theorem: $(1+i)^n = [\sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})]^n = 2^{\frac{n}{2}}(\cos \frac{n \pi}{4} + i \sin \frac{n \pi}{4})$. Equating the real parts: $a_0 - a_2 + a_4 - a_6 + \ldots = 2^{\frac{n}{2}} \cos \frac{n \pi}{4}$. Comparing this with the given expression $k \cos \frac{n \pi}{4}$,we get $k = 2^{\frac{n}{2}}$. Thus,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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