Prove $7^{2n}-1$ is divisible by 48 Trying to prove $7^{2n}-1$ is divisible by 48.
I began by taking the base case $n=1$ so $7^{2}-1=48$ which is true.
Now from here taking the inductive step to get
$7^{2(n+1)}-1=48k$
$7^{2n+2}-1=48k$
$7^{2}*7^{2n}-1=48k$
Not quite sure where to go from here. Thanks.
 A: First, it is bad practice to write the proofs in this way, beginning with what you want to show and rewriting until you get a true statement. This is very simply backwards: a proof should always go from what we know (or have assumed) to our desired conclusion.
So, your inductive step would be much better written as a string of equalities, i.e.
$$7^{2(n+1)} - 1 = 7^{2n+2} - 1 = 7^2 * 7^{2n} - 1 = \text{...[fill this in]}.$$
Beyond this, your inductive step should always begin by stating your inductive hypothesis! So overall, your inductive step should look like:

Suppose that, for some positive integer $n$, there is an integer $k$ such that $7^{2n} - 1 = 48k$. Then
$$7^{2(n+1)} - 1 = 7^{2n+2} - 1 = 7^2 * 7^{2n} - 1 = \text{...[fill this in]}.$$

Now it's a little more clear what to do: apply the inductive hypothesis to rewrite $7^{2n}$ as $48k+1$! Hope this helps.
A: Let S = $1 + 49 + 49^2 + \cdots + 49^{n-1} $
\begin{align}
        S &= 1 + 49 + 49^2 + \cdots + 49^{n-1} + 49^n - 49^n\\
        S &= 1 + 49(1 + 49 + 49^2 + \cdots + 49^{n-1}) - 49^n \\
        S &= 1 + 49S - 49^n \\
 49^n - 1 &= 48S\\
       48 &\mid 49^n - 1 \\
       48 &\mid 7^{2n} - 1 \\
\end{align}
A: You can use
$7^{2n}-1=(1+48)^n-1$
Using binomial theorem
$7^{2n}-1= 1+48n + \frac{n(n-1)}{2!}48^2 + \frac{n(n-1)(n-2)}{3!} 48^3+....+ \frac{n!}{n!}48^n -1$
from here you can see
$48|(7^{2n}-1)$
A: It does not need induction. Just see that
$$7^{2n} = 49^n \equiv 1^n = 1\mod 48$$
and therefore $$48 \mid 7^{2n}-1.$$
But, if you really want an inductive proof, let $a_n=7^{2n}-1$.
$a_1=48$ and then $48\mid a_1$. Notice that
$$a_{n+1} = 7^{2n+2}-1 = 49\cdot 7^{2n}-1 = 49 (7^{2n}-1) + 48 = 49 a_n + 48.$$
Suppose that $48\mid a_n$, that is, $a_n= 48k$ for some $k\in\mathbb{Z}$. Then
$$a_{n+1} = 49 a_n + 48 = 49 \cdot 48 k + 48 = 48 (49k+1) $$
and therefore $48\mid a_{n+1}$. The results follows by the principle of Induction.
