If (1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + | vedclass

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(D) Given $(1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + C_nx^n$. Put $x = i$ on both sides: $(1 + i)^n = (C_0 - C_2 + C_4 - .....) + i(C_1 - C_3 + C_5 -

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$2^{n/2} \cos \frac{n\pi}{4}$

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(D) Given $(1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + C_nx^n$. Put $x = i$ on both sides: $(1 + i)^n = (C_0 - C_2 + C_4 - .....) + i(C_1 - C_3 + C_5 - .....) \dots (i)$ Express $1 + i$ in polar form: $1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$. Using De Moivre's theorem: $(1 + i)^n = (\sqrt{2})^n \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)^n = 2^{n/2} \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) \dots (ii)$ Equating the real parts from $(i)$ and $(ii)$,we get: $C_0 - C_2 + C_4 - C_6 + ..... = 2^{n/2} \cos \frac{n\pi}{4}$.

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + C_nx^n$,then the value of $C_0 - C_2 + C_4 - C_6 + .....$ is

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