Recursive sequence

$a_{n+1} = 3 a_{n} +2$

$\sum_{i=1}^{n} a_{i+1} = 3 \sum_{i=1}^{n} a_{i} + 2n$

$a_{n+1} + \sum_{i=1}^{n} a_{i} - a_{1} =3 \sum_{i=1}^{n} a_{i} + 2n$

Let $S= \sum_{i=1}^{100} a_{i}$ and substitute $n=100$ .

So, $a_{101} + S -a_{1} = 3S + 200$

$3a_{100} + 2 -2 = 2S +200$

$2S + 200 = 3 a_{100}$

Therefore, $\frac{2 \sum_{i=1}^{n} a_{i} +200}{a_{100}} = \boxed{3}$

$a_{n+1}=3a_{n}+2$

$a_{n+1}+1=3a_{n}+3=3(a_{n}+1)$

Let $b_{n}=a_{n}+1$ then $b_{n+1}=3b_{n}$ and $b_{1}=3$ so $b_{n}=3^{n}$

Now we see $a_{n}=3^{n}-1$

$\frac{[2\sum\limits_{n=1}^{100}(3^{n}-1)]+200}{3^{100}-1}=\frac{[2\frac{3(3^{100}-1)}{2}-200]+200}{3^{100}-1}=\boxed{3}$