If C_r = { }^n C_r,then find the sum C_0 + C_ | vedclass

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(A) Let $S = C_0 + C_4 + C_8 + C_{12} + \ldots$ Consider the expansion $(1+x)^n = \sum_{r=0}^n C_r x^r$. Let $\omega$ be a fourth root of unity,i.e.,$

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$\frac{2^{\frac{n}{2}} \left[ \cos \frac{n \pi}{4} + 2^{\frac{n}{2}-1} \right]}{2}$

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(A) Let $S = C_0 + C_4 + C_8 + C_{12} + \ldots$ Consider the expansion $(1+x)^n = \sum_{r=0}^n C_r x^r$. Let $\omega$ be a fourth root of unity,i.e.,$\omega = i$. The roots are $1, i, -1, -i$. Using the property of roots of unity,$\sum_{k=0}^3 (1 + \omega^k x)^n = \sum_{k=0}^3 \sum_{r=0}^n C_r (\omega^k)^r x^r = \sum_{r=0}^n C_r x^r \sum_{k=0}^3 (\omega^r)^k$. The inner sum $\sum_{k=0}^3 (\omega^r)^k$ is $4$ if $r$ is a multiple of $4$ (i.e.,$r = 0, 4, 8, \ldots$) and $0$ otherwise. Thus,$4S = (1+1)^n + (1+i)^n + (1-1)^n + (1-i)^n = 2^n + (1+i)^n + 0^n + (1-i)^n$. Since $(1+i) = \sqrt{2} e^{i \pi/4}$ and $(1-i) = \sqrt{2} e^{-i \pi/4}$,we have: $(1+i)^n + (1-i)^n = 2^{n/2} (e^{i n \pi/4} + e^{-i n \pi/4}) = 2^{n/2} \cdot 2 \cos \frac{n \pi}{4} = 2^{n/2+1} \cos \frac{n \pi}{4}$. Therefore,$4S = 2^n + 2^{n/2+1} \cos \frac{n \pi}{4}$. $S = \frac{2^n}{4} + \frac{2^{n/2+1}}{4} \cos \frac{n \pi}{4} = 2^{n-2} + 2^{n/2-1} \cos \frac{n \pi}{4}$. This can be rewritten as $\frac{2^{n/2} [\cos \frac{n \pi}{4} + 2^{n/2-1}]}{2}$.

If $C_r = { }^n C_r$,then find the sum $C_0 + C_4 + C_8 + C_{12} + \ldots$

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