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

Question

Aicharya · Accepted Answer

In given series difference of suffies is 4 $ 0-4=8-4=12-4=\ldots=4 $ Now, $(\mathrm{l})^{\frac{1}{4}}=(\cos 0+i \sin 0)^{\frac{1}{4}}$ $ \begin{aligned} & =(\cos 2 r \pi+i \sin 2 r \pi)^{\frac{1}{4}} \\ & =\cos \frac{r \pi}{2}+i \sin \frac{r \pi}{2} \end{aligned} $ where $r=0,1,2,3$ Four roots of unity are $1, i,-1,-i$ $ \begin{aligned} & =1, \alpha, \alpha, \alpha^2 \quad \text{say}\\ \text { and }(1+x)^n & =\sum_{r=0}^n{ }^n C_r x^r \end{aligned} $

For $r=0,4,8,12 \ldots$ R.H.S of Eq. (v) $ \begin{array}{r} { }^n C_0(1+1+1+1)+{ }^n C_4\left(1+\alpha^4+\alpha^8+\alpha^{12}\right) \\ +{ }^n C_8\left(1+\alpha^8+\alpha^{16}+\alpha^{24}\right)+\ldots+ \\ =4\left({ }^n C_0+{ }^n C_4+{ }^n C_8 \ldots\right) \quad\left(\because \alpha^4=1\right) . \end{array} $ L.H.S of Eq. (v) $ \begin{aligned} & =2^n+(1+i)^n+\left(1+i^2\right)^n+\left(1+i^3\right)^n \\ & =2^n+(1+i)^n+0+(1-i)^n \end{aligned} $ Since, $(1+i)^n=2^{\frac{n}{2}}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^n$ $ =2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right] $ $ \begin{aligned} & \quad(1-i)^n=2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}-i \sin \frac{n \pi}{4}\right] \\ & \Rightarrow 2^n+(1+i)^n+(1-i)^n \\ & =2^n+2 \cdot 2^{\frac{n}{2}} \cos \frac{n \pi}{4} \\ & \Rightarrow 4 \cdot\left({ }^n C_0+{ }^n C_4+{ }^n C_8+{ }^n C_{12} \cdots\right) \\ & \quad=2 \cdot 2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]=\frac{2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]}{2} \end{aligned} $

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

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