If C_0, C_1, C_2, ldots, C_{n} are the binomi | vedclass

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(D) Let the given expression be $S = (C_0+C_1)-(C_2+C_3)+(C_4+C_5)-(C_6+C_7)+\ldots$ This can be written as $S = (C_0-C_2+C_4-C_6+\ldots) + (C_1-C_3+C

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$2^{n/2} \left(\cos \frac{n\pi}{4} + \sin \frac{n\pi}{4}\right)$

Vedclass · Answer

(D) Let the given expression be $S = (C_0+C_1)-(C_2+C_3)+(C_4+C_5)-(C_6+C_7)+\ldots$ This can be written as $S = (C_0-C_2+C_4-C_6+\ldots) + (C_1-C_3+C_5-C_7+\ldots)$. Consider the expansion $(1+i)^n = C_0 + C_1 i + C_2 i^2 + C_3 i^3 + C_4 i^4 + C_5 i^5 + C_6 i^6 + C_7 i^7 + \ldots$ Since $i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, i^6 = -1, i^7 = -i$,we have: $(1+i)^n = (C_0 - C_2 + C_4 - C_6 + \ldots) + i(C_1 - C_3 + C_5 - C_7 + \ldots)$. Let $A = (C_0 - C_2 + C_4 - C_6 + \ldots)$ and $B = (C_1 - C_3 + C_5 - C_7 + \ldots)$. Then $(1+i)^n = A + iB$. The given expression is $S = A + B$. We know $1+i = \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)$. So,$(1+i)^n = (\sqrt{2})^n \left(\cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4}\right) = 2^{n/2} \left(\cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4}\right)$. Thus,$A = 2^{n/2} \cos \frac{n\pi}{4}$ and $B = 2^{n/2} \sin \frac{n\pi}{4}$. Therefore,$S = A + B = 2^{n/2} \left(\cos \frac{n\pi}{4} + \sin \frac{n\pi}{4}\right)$.

If $C_0, C_1, C_2, \ldots, C_{n}$ are the binomial coefficients in the expansion of $(1+x)^{n}$,then $(C_0+C_1)-(C_2+C_3)+(C_4+C_5)-(C_6+C_7)+\ldots=$

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