Compare two halves of a binomial expansion

$\begin{aligned} B & ={\color{#3D99F6}\sum_{i=t+1}^{2t+1} \binom {2t+1}i} 2019^i & \small \color{#3D99F6} \text{Note that }\sum_{i=t+1}^{2t+1} \binom {2t+1}i = \sum_{i=0}^t \binom {2t+1}i \\ & = {\color{#3D99F6} \sum_{i=0}^t \binom {2t+1}i} 2019^{2t+1-i} \\ & = {\color{#3D99F6} \sum_{i=0}^t \binom {2t+1}i} 2019^{i + \color{#D61F06}(2t-2i+1)} & \small \color{#D61F06} \text{Since }2t - 2i + 1 \ge 1 \\ & \boxed{> 2019A} \end{aligned}$

Have a closer look at $B$ and define a new counting variable: $j = n - i$ . Then $j$ is from $n - (t+1) = 2t+1 - (t+1) = t$ to $n - n = 0$ , while ${n \choose i} = \frac{n!}{i!(n-i)!} = {n \choose (n - i)} = {n \choose j}$ . So $B$ can be rearranged and written as: $B = \sum_{j = 0}^{t}{n \choose j}2019^{n-j}$ .

Consider $B - 2019A$ we have: $B - 2019A = \sum_{j = 0}^{t}{n \choose j}2019^{n-j} - 2019\sum_{i = 0}^{t}{n \choose i}2019^i\ = \sum_{j = 0}^{t}{n \choose j}2019^{2t+1-j} - \sum_{j = 0}^{t}{n \choose j}2019^{j+1}$ (rename the variable in the latter sum from $i$ to $j$ ) $=\sum_{j = 0}^{t}{n \choose j}(2019^{2t-j+1} - 2019^{j+1})$ .

For any $j$ from $0$ to $t - 1$ we see that $2019^{2t - j +1} - 2019^{j +1} > 0$ (note the case $j = 0$ ) while at $j = t$ , $2019^{2t- j +1} - 2019^{j+1} = 2019^{t+1} - 2019^{t+1} = 0$ . This ensures ${n \choose j}(2019^{2t - j} - 2019^j) \geq 0$ for all $j \in \{0,1,...,t\}$ . And because the term $j = 0$ is present regardless of $t$ , $B - 2019A = \sum_{j = 0}^{t}{n \choose j}(2019^{2t - j+1} - 2019^{j+1}) > 0$ for all positive integer $t$ , implying $B > 2019 A$ .

$t=1\Longrightarrow 1$ ; $n=2 t+1 \Longrightarrow 3$ ; assuming $x\geq 1$ , which it is as $x=2019$ , $(\sum _{i=0}^t x^{i+1} \binom{n}{i} \Longrightarrow 3\ x^2+x)<(\sum _{i=t+1}^n x^i \binom{n}{i} \Longrightarrow x^3+3 x^2)$ . The difference only becomes greater as $t$ becomes larger.

Wolfram, using t=1 meaning n=3
You can see that B is much larger than 2019*A