Proof by induction $3^{2n}-2^n$ is multiply of 7

Question

Proof $3^{2n}-2^n$ is multiply of 7

First $n=1$, $9-2=7$

For N+1, $3^{2(n+1)}-2^{n+1}=3^{2n+2}-2^n\cdot2$=$9\cdot3^{2n}-2\cdot2^n$

Can someone give me hint how can i get multiply of 7?

Answer 1

For the induction step note that assuming as hypotesis

$$3^{2{n}}- 2^{n}=7k$$

we have

$$3^{2{(n+1)}}-2^{n+1}=9\cdot3^{2{n}}-2\cdot 2^{n}=9\cdot 7k +9\cdot2^{{n}}-2\cdot 2^{n}=7(9k+2^n)$$

Answer 2

$3^{2(n+1)}-2^{n+1}=9\cdot 3^{2n}-2 \cdot2^n=2(3^{2n}-2^n)+7 \cdot 3^{2n}$.

Answer 3

To continue the inductive step notice that;

$9\cdot 3^{2n}-2\cdot 2^n = 9\cdot 3^{2n}-(9-7)\cdot 2^n= 9(3^{2n}-2^n)+7\cdot 2^n$ ,

which is divisible by $7$