A Nice Problem!

Let $I$ denote the summation. Apply the property of binomial coefficient: $\dbinom{n}{a} = \dbinom{n}{n-a}$ .

$\begin{aligned} \displaystyle I + I &=& \sum_{n=0}^7 \left(\dbinom{30}{4n} + \dbinom{30}{30-4n}\right) \\ \displaystyle 2I &=& \sum_{n \text{ even}} \dbinom{30}{n} \\ 2I & = & \sum_k { 29 \choose k } & \text{ since } { 30 \choose n} = {29 \choose n-1} + { 29 \choose n } \\ \displaystyle 2I &=& \cdot 2^{29} \\ I &=& 2^{28} \\ \end{aligned}$

$\begin{aligned} (1+i)^{30} & = \small \sum_{n=0}^{30} \begin{pmatrix} 30 \\ n \end{pmatrix} i^n \\ \left(\sqrt{2}e^{\frac{\pi}{4}i}\right)^{30} & = 1 + 30i - 435 - 4060i +...-4060i+435+30i-1 \\ 2^{15} e^{\frac{15\pi}{2}i} & = \small \sum_{n=0}^7 \begin{pmatrix} 30 \\ 4n \end{pmatrix} + i \sum_{n=0}^7 \begin{pmatrix} 30 \\ 4n+1 \end{pmatrix} - \sum_{n=0}^7 \begin{pmatrix} 30 \\ 4n+2 \end{pmatrix} - i \sum_{n=0}^6 \begin{pmatrix} 30 \\ 4n+3 \end{pmatrix} \\ 0 - 2^{15} i & = S_0 + S_1i - S_2 - S_3i \\ \color{#3D99F6}{\Rightarrow S_0} & = \color{#3D99F6}{S_2} \end{aligned}$

Now, we have:

$\begin{cases} (1+1)^{30} = S_0 + S_1 + S_2 + S_3 = 2^{30} &...(1) \\ (1-1)^{30} = S_0 - S_1 + S_2 - S_3 = 0 &...(2) \end{cases}$

$\begin{aligned} (1)+(2): \quad 2S_0 + 2S_2 & = 2^{30} \\ 4S_0 & = 2^{30} \\ \Rightarrow S_0 & = \color{#3D99F6} {\sum_{n=0}^7 \begin{pmatrix} 30 \\ 4n \end{pmatrix} = \boxed{2^{28}}} \end{aligned}$

The sum was relatively small so I was able to use a calculator to obtain a = 28. However, when I tried the complex approach all I got was $\ (1+i)^{30} = i ( [ {30 \choose 1 } +{30 \choose 5 } +...+ {30 \choose 29 } ] - [{30 \choose 3 } +{30 \choose 7 }+...+{30 \choose 27 } ])$ $\ = 2^{15} i$ I also tried the same for (1-i) but ended up with the same result. What am I missing?