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 $k+1$.
 A: Assuming that $7|(3^{2n-1} + 2^{n+1})$ then
$$3^{2n+1} + 2^{n+2}=3^2 \cdot 3^{2n-1} +2\cdot 2^{n+1}\equiv 2\cdot (3^{2n-1} + 2^{n+1}) \equiv 0\mod 7$$
which proves the induction step.
A: 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,
$$3^{2n-1} + 2^{n+1}= 3\cdot 9^{n-1} + 4\cdot 2^{n-1}$$
$$= 3\cdot 9^{n-1} + (7-3)\cdot 2^{n-1} = 3(\underbrace{9^{n-1} - 2^{n-1}}_{\stackrel{ (\star)}{=}7\cdot m}) + 7\cdot 2^{n-1}$$
A: $3^{2n-1}  \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &3 & 6 & 5 & 3 & 6 & \cdots \end{matrix}$
$2^{n+1} \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &4 & 1 & 2 & 4 & 1 & \cdots \end{matrix}$
Since $3+4 = 7$, $6+1=7$, $5+2 = 7$ we can see that $3^{2n-1}+2^{n+1} \equiv 0 \pmod{7}$
A: $(3^2)^{n-1}+2^{n+1}=$
$ (1/3)(7+2)^n +2 \cdot 2^n=$
$(1/3)(\sum_{k=0}^{n}\binom {n}{k} 7^{n-k}2^k+6\cdot 2^n)=$
$(1/3)(\sum_{k=0}^{n-1} \binom{n}{k} 7^{n-k} 2^k +7 \cdot 2^n)$;
All terms in the sum above have a factor $7$.
A: Hint: Write $$(3^2)^n\times3^{-1}+2^n\times 2$$ and $$3^2\equiv 2\mod n$$
A: Hint:
If $f(n)=3^{2n-1}+2^{n+1}$
$$f(m+1)-2f(m)=3^{2m-1}(3^2-2)$$ which is obviously divisible by $7$
So, if $7|f(m),7|f(m+1)$
If induction is not mandatory,
$$3^{2n-1}+2^{n+1}=3(3^2)^{n-1}+2^{n+1}\equiv3\cdot2^{n-1}+2^{n+1}\pmod7\equiv2^{n-1}(3+2^2)$$
