Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$ I have tried to prove this through mathematical induction but I can't seem to prove that the proposition works for $k+1$.

You can extract the factor $7$ as follows using


*

*$(\star)$: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots +  ab^{n-2} + b^{n-1})$
Hence,
$$3^{2n-1} + 2^{n+1}= 3\cdot 9^{n-1} + 4\cdot 2^{n-1}$$
$$= 3\cdot 9^{n-1} + (7-3)\cdot 2^{n-1} = 3(\underbrace{9^{n-1} - 2^{n-1}}_{\stackrel{ (\star)}{=}7\cdot m}) + 7\cdot 2^{n-1}$$
