Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$ 
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \times 2 = 7 \times 3^{2k+1} + 2 \times 3^{2k+1} + 2^{k-1} \times 2.$$
Where do I go from here?

Note that from your assumption you have:
$3^{2k+1} +2^{k-1} = 7p,\quad p\in\mathbb{Z}$
Then you can change the following line you have:
$9 \cdot3^{2k+1} + 2^{k-1} \cdot 2$
into:
$2(3^{2k+1}+2^{k-1})+7\cdot3^{2k+1} = 2\cdot 7p+7\cdot3^{2k+1}=7(2p+3^{2k+1})$
which is divisible by $7$ as required.
