, 2018

Solution:

Let $s_{n}$ be the sum of the absolute values of numbers in the $n$th row. For odd $n$, we have that $a_{n, 1}, \ldots, a_{n, n-1}$ alternate in sign as $-,+,-,+, \ldots,+$, with the last term being $a_{n, n-1}=1$. For even $n$, we have that $a_{n, 1}, \ldots, a_{n, n-2}$ alternate in sign as $-,+,-,+, \ldots,+$, and $a_{n, n-1}=0$. These facts can be proven by induction. Thus, $s_{n}=1-a_{n, 1}+a_{n, 2}-\cdots+(-1)^{n-1} a_{n, n-1}+1$.

Applying the recursion, for $n>0$ this becomes
$$s_{n}=1-\left(a_{n-1,1}-a_{n-1,0}\right)+\left(a_{n-1,2}-a_{n-1,1}\right)-\cdots+(-1)^{n-1}\left(a_{n-1, n-1}-a_{n-1, n-2}\right)+1$$
$$=2\left(1-a_{n-1,1}+a_{n-1,2}-\cdots+(-1)^{n-2} a_{n-1, n-2}+1\right)-1+(-1)^{n-1}$$
In other words, if $n$ is even then $s_{n}=2 s_{n-1}-2$ and if $n$ is odd then $s_{n}=2 s_{n-1}$. This means that $s_{2 n}=4 s_{2 n-2}-2$. Since 2018 is even, we can write $s_{2018}=4 s_{2016}-2=2^{2018}-2^{2017}-2^{2015}-\cdots-2$.

Applying the formula for the sum of a geometric series, we get
$$s_{2018}=2^{2018}-\frac{2^{2019}-2}{4-1}=\frac{2^{2018}+2}{3}$$