Complex number #1

Note that $\displaystyle \sum_{k=0}^{\color{#3D99F6}\infty} \binom nk = \sum_{k=0}^{\color{#D61F06}n} \binom nk$ , since $\displaystyle \binom nk = 0$ for $k>n$ . By binomial theorem , we have $\displaystyle (1+x)^n = \sum_{k=0}^n \binom nk x^k$ , therefore:

$\begin{aligned} (1+x)^n & = 1 + \binom n1 x + \binom n2 x^2 + \binom n3 x^3 + \binom n4 x^4 + \binom n5 x^5+ \cdots + \binom nn x^n \\ \frac {(1+x)^n -1}x & = \binom n1 + \binom n2 x + \binom n3 x^2 + \binom n4 x^3 + \binom n5 x^4+ \cdots + \binom nn x^{n-1} \\ \frac {(1+{\color{#3D99F6}1})^n -1}{\color{#3D99F6}1} & = \binom n1{\color{#3D99F6}(1)} + \binom n2{\color{#3D99F6}(1)} + \binom n3{\color{#3D99F6}(1)} + \binom n4{\color{#3D99F6}(1)} + \binom n5 {\color{#3D99F6}(1)} + \cdots + \binom nn {\color{#3D99F6}(1)} \\ \frac {(1+{\color{#3D99F6}i})^n -1}{\color{#3D99F6}i} & = \binom n1{\color{#3D99F6}(1)} + \binom n2{\color{#3D99F6}(i)} + \binom n3{\color{#3D99F6}(-1)} + \binom n4{\color{#3D99F6}(-i)} + \binom n5 {\color{#3D99F6}(1)} + \cdots + \binom nn {\color{#3D99F6}(i)} \\ \frac {(1\ {\color{#3D99F6}-1})^n -1}{\color{#3D99F6}-1} & = \binom n1{\color{#3D99F6}(1)} + \binom n2{\color{#3D99F6}(-1)} + \binom n3{\color{#3D99F6}(1)} + \binom n4{\color{#3D99F6}(-1)} + \binom n5 {\color{#3D99F6}(1)} + \cdots + \binom nn {\color{#3D99F6}(-1)} \\ \frac {(1\ {\color{#3D99F6}-i})^n -1}{\color{#3D99F6}-i} & = \binom n1{\color{#3D99F6}(1)} + \binom n2{\color{#3D99F6}(-i)} + \binom n3{\color{#3D99F6}(-1)} + \binom n4{\color{#3D99F6}(i)} + \binom n5 {\color{#3D99F6}(1)} + \cdots + \binom nn {\color{#3D99F6}(-i)} \end{aligned}$

This implies that:

$\begin{aligned} \frac {(1+1)^n -1}1 + \frac {(1+i)^n -1}i + \frac {(1-1)^n -1}{-1} + \frac {(1-i)^n -1}{-i} & = 4 \binom n1 + 4 \binom n5 + 4 \binom n9 + 4 \binom n{13} + \cdots + 4 \binom n{n-1} & \small \color{#3D99F6} \text{Put }n=2018 \end{aligned}$

$\begin{aligned} \implies \sum_{k=1}^\infty \binom {2018}{4k+1} & = \frac {2^{2018} - 1 -i(1+i)^{2018} + i - 0 + 1 + i(1-i)^{2018} - i}4 & \small \color{#3D99F6} \text{Note that }(1+i)^2 = 2i, \ (1-i)^2 = -2i \\ & = \frac {2^{2018} - i(2i)^{1009} + i(-2i)^{1009}}4 \\ & = \frac {2^{2018} - 2^{1009}i^{1010} - 2^{1009}i^{1010}}4 & \small \color{#3D99F6} \text{Note that }i^4 = 1, i^2 = -1 \\ & = \frac {2^{2018} + 2^{1009} + 2^{1009}}4 \\ & = \frac {2^{2018} + 2^{1010}}4 \\ & = 2^{2016} + 2^{1008} \\ & = 2^{1008}\left(2^{1008}+2\right) \end{aligned}$

Therefore, $a = \boxed{1008}$ .

Consider $f(x) = x^3 (1 + x)^{2018}$ . Then our answer is $\frac{f(1) + f(i) + f(-i) + f(-1)}{4} = 2^{1008}(2^{1008} + 1) .$