Unit digit from $A$

$\begin{aligned} A & = \frac {\sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}2017}\binom {2018}k}2+1 \\ & = \frac {\sum_{\color{#D61F06}k=0}^{\color{#D61F06}2018}\binom {2018}k - \binom {2018}0 - \binom {2018}{2018}}2+1 \\ & = \frac {(1+1)^{2018} - 1 - 1}2+1 \\ & = 2^{2017} - 1 +1 \\ & = 2^{2017} \end{aligned}$

We need to find $2^{2017} \bmod 10$ . Since $\gcd (2,10) \ne 1$ , we have to consider the factors 2 and 5 of 10 separately using the Chinese remainder theorem .

Consider factor 2: $A \equiv 2^{2017} \equiv 0 \text{ (mod 2)}$

Consider factor 5:

$\begin{aligned} A & \equiv 2^{\color{#3D99F6}2017 \bmod \phi(5)} \text{ (mod 5)} & \small \color{#3D99F6} \text{Since }\gcd(2,5) = 1\text{, Euler's theorem applies.} \\ & \equiv 2^{\color{#3D99F6}2017 \bmod 4} \text{ (mod 5)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (5) = 4 \\ & \equiv 2 \text{ (mod 5)} \\ \implies A & \equiv 5n + 2 & \small \color{#3D99F6} \text{where }n \text{ is an integer.} \end{aligned}$

Then, we have:

$\begin{aligned} 5n+2 & \equiv 0 \text{ (mod 2)} \\ \implies n & \equiv 0 \\ \implies A & \equiv 5(0) + 2 \equiv \boxed{2} \text{ (mod 10)} \end{aligned}$

$\displaystyle \sum_{k=1}^{2017}\binom{2018}{k} = \binom{2018}{1} + \binom {2018}{2} + \binom{2018}{3} + ... + \binom{2018}{2017} \\ \displaystyle \sum_{k=1}^{2017}\binom{2018}{k} = \binom{2018}{0} + \binom{2018}{1} + \binom {2018}{2} + \binom{2018}{3} + ... + \binom{2018}{2017} + \binom{2018}{2018} - \binom{2018}{0} - \binom{2018}{2018} \\ \displaystyle \sum_{k=1}^{2017}\binom{2018}{k}= 2^{2018} - 1 - 1 \\ \displaystyle \therefore \sum_{k=1}^{2017}\binom{2018}{k}= 2^{2018} - 2$

So, we get

$\frac{\displaystyle \sum_{k=1}^{2017}\binom{2018}{k}}{2} + 1 = \frac{2^{2018} - 2}{2} + 1 \\ \frac{\displaystyle \sum_{k=1}^{2017}\binom{2018}{k}}{2} + 1 = \frac{2(2^{2017} - 1)}{2} + 1 \\ \frac{\displaystyle \sum_{k=1}^{2017}\binom{2018}{k}}{2} + 1 = 2^{2017} - 1 + 1 \\ \therefore \frac{\displaystyle \sum_{k=1}^{2017}\binom{2018}{k}}{2} + 1 = 2^{2017}$

$2^1 \equiv 2 \mod 10 \\ 2^2 \equiv 4 \mod 10 \\ 2^3 \equiv 8 \mod 10 \\ 2^4 \equiv 6 \mod 10 \\ 2^5 \equiv 2 \mod 10$

It appears that the pattern repeats after 4 patterns. So, $2017 \equiv 1 \mod 4$ . Then the unit number of $2 ^ {2017}$ is the same as the unit number of $2 ^ {1}$ , is $2$ .

$\sum_{k=1}^{2017} \binom{2018}k = \left(\sum_{k = 0}^{2018}\ \binom{2018}k\right) - 2 = 2^{2018} - 2.$ Therefore $A = \frac{2^{2018} - 2}2 + 1 = 2^{2017} - 1 + 1 = 2^{2017}.$ Now the unit digit $2^n$ is $\begin{cases} 1 & n = 0 \\ 2 & n \equiv 1\ \text{mod}\ 4 \\ 4 & n \equiv 2\ \text{mod}\ 4 \\ 8 & n \equiv 3\ \text{mod}\ 4 \\ 6 & n \equiv 0\ \text{mod}\ 4\ \text{and}\ n > 0 \end{cases}$ Since $2017 \equiv 1$ mod 4, the unit digit is equal to $\boxed{2}$ .