$7^n + 2\cdot 4^n$ is divisible by $3$ (Proof by induction)

Question

Can someone please help? I'm stuck with this question.

I've currently got if $P(k)$ is true then $7^k +2 \cdot 4^k = 3A$ for all $n \in \mathbb Z^+$

Now $7^{k+1} + 2 \cdot (4^{k+1}) = 7 \cdot 7^k + 2 \cdot 4\cdot4^k$

Answer 1

Notice: $7^n+2\times 4^n=7^n+2^{2n+1}$.

Now suppose $7^n+2^{2n+1}=3q$, then $$7^{n+1}+2^{2(n+1)+1}=7^{n+1}+2^{2n+3}$$ $$=7(7^n)+4(2^{2n+1})=7(7^n)+7(2^{2n+1})-3(2^{2n+1})$$ $$=7(7^n+2^{2n+1})-3(2^{2n+1})=7(3q)-3(2^{2n+1})=3(7q-2^{2n+1}) $$ and we are done.

Edit: $3$ divides $7^n+2^{2n+1}$, i.e. there exists an integer $q$ such that $7^n+2^{2n+1}=3q$.

Now to prove that $3$ devises$7^{n+1}+2^{2n+3}$, we need to find an integer $q'$ such that $7^{n+1}+2^{2n+3}=3q'$ and the desired $q'$ is $7q-2^{2n+1}$.

Answer 2

Hint:

Let $P(n)=7^n+2\cdot4^n$.

Then $P(n+1)-P(n)=?$

Answer 3

If you want, have another way to solve this. Notice that,$ 7 \equiv 1 \pmod 3 \Rightarrow 7^n \equiv 1 \pmod3$ , $ 4 \equiv 1 \pmod 3 \Rightarrow 4^n \equiv 1 \pmod3$. Then $7^n+ 2 \cdot 4^n \equiv 1+2=3 \equiv 0 \pmod3$. Therefore, $3 \mid 7^n + 2 \cdot 4^n$