Prove by mathematical induction that Prove by mathematical induction that 
$$P(n)=3^{2n+1} + 2^{n-1}$$
is a multiple of 7.
My Attempt:
Here, $P(n)= 3^{2n+1}+2^{n-1}$
For $n=1$,
$$P(1)=3^3+2^0$$
$$=28$$
So, $P(1)$ is true. 
Suppose, $P(m)$ is true for all $m\in N$
Now, we have to show that $P(m+1)$ is true,
$$P(m+1)=3^{2(m+1)+1}+2^{(m+1)-1}$$
$$=3^{2m+3}+2^m$$
 A: Inductive Step
$$
\begin{align}
P(m+1)
&=3^{2m+3}+2^m\\
&=9\cdot3^{2m+1}+2\cdot2^{m-1}\\
&=7\cdot3^{2m+1}+2\left(3^{2m+1}+2^{m-1}\right)\\
&=7\cdot3^{2m+1}+2P(m)
\end{align}
$$

Direct Proof
Note that since $2\cdot4\equiv1\pmod7$, we get
$$
\begin{align}
3^{2k+1}+2^{k-1}
&=3\cdot9^k+2^{-1}\cdot2^k\\
&\equiv3\cdot2^k+4\cdot2^k&\pmod7\\
&=7\cdot2^k\\
&\equiv0&\pmod7
\end{align}
$$
A: it is $$P(n+1)=3^{2n+3}+2^n$$ and we get $$P(n+1)-P(n)=3^{2n+1}\cdot 7+3^{2n+1}+2^{n-1}$$ therefore $$7|P(n+1)$$
A: 
Now, we have to show that $P(m+1)$ is true,
  $$P(m+1)=3^{2(m+1)+1}+2^{(m+1)-1}$$
  $$=3^{2m+3}+2^m$$

You'll want to use the hypothesis of the inductive step ("$P(m)$ is true"); so rewrite:
$$\begin{align}
3^{2m+3}+2^m 
& =3^2 \cdot 3^{2m+1}\color{red}{+2}\cdot 2^{m-1} \\[5pt]
& =9 \cdot 3^{2m+1}\color{red}{+9}\cdot 2^{m-1}\color{red}{-7}\cdot 2^{m-1} \\[5pt]
& = 9 \left(\color{blue}{3^{2m+1}+2^{m-1}} \right) - 7\cdot 2^{m-1}
\end{align}$$
Now the first term is divisible by $7$ because (...) and the second term is (...).
A: Now, $$3^{2m+3}+2^m=9\cdot3^{2m+1}+2^m=9\left(3^{2m+1}+2^{m-1}-2^{m-1}\right)+2\cdot2^{m-1}=$$
$$=9\left(3^{2m+1}+2^{m-1}\right)-9\cdot2^{m-1}+2\cdot2^{m-1}=9\left(3^{2m+1}+2^{m-1}\right)-7\cdot2^{m-1}.$$
Can you end it now?
A: There are several standard ways to cope with the problem. First, state $P(n)$ properly:

$P(n)$: for every $n\ge1$, there exists $k$ such that $3^{2n+1}+2^{n-1}=7k$

(variables are assumed to vary in the natural numbers).
$P(1)$: $3^{2+1}+2^{1-1}=3^3+1=28=7\cdot 4$
Therefore $P(1)$ is true.
Assume $P(n)$, that is, $3^{2n+1}=7k-2^{n-1}$ for some $k$. Then
$$
3^{2(n+1)+1}+2^{(n+1)-1}=
3^2\cdot 3^{2n+1}+2^n=
9\cdot 7k-9\cdot2^{n-1}+2\cdot2^{n-1}=7(9k-2^{n-1})
$$
Therefore we have proved that, for every $n\ge1$, $P(n)\implies P(n+1)$.
