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?

By the binomial theorem
$$
3^{2n+1}=3\cdot 9^n=3(7+2)^n=3(7a+2^n)=21a+6\cdot2^{n-1}
$$
Therefore
$$
3^{2n+1}+2^{n-1}=21a+7\cdot2^{n-1}=7(3a+2^{n-1})
$$
