Recursive Sequence Term

$a_1 = 1$ $a_2 = 2(1) + 5 = 7$ $a_3 = 2(7) + 5 = 19$ $a_4 = 2(19) + 5 = 43$

The differences are $6, 12, 24, \dots$ which is a geometric progression with first term $6$ and common ratio $2$. $a_n = a_1 + \sum_{k=1}^{n-1} 6 \cdot 2^{k-1} = 1 + 6(2^{n-1} - 1) = 6 \cdot 2^{n-1} - 5$. For $n = 100$, $a_{100} = 6 \times 2^{99} - 5$.