Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use. 

${\displaystyle{\frac{1}{9}}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Proof by induction: 
For $n=1, {\displaystyle{\frac{1}{9}}(10^1+3 \cdot 4^1 + 5) = \frac{27}{9} = 3}$, so the result holds for $n=1$
Assume the result to be true for $n=m$, i.e.  $\displaystyle{\frac{1}{9}(10^m+3 \cdot 4^m + 5)}$ is an integer
To show ${\displaystyle{\frac{1}{9}}(10^{m+1}+3 \cdot 4^{m+1} + 5)}$ is an integer.
$$
\begin{align*}
\displaystyle{\frac{1}{9}(10^{m+1}+3 \cdot 4^{m+1} + 5) -(10^m+3 \cdot 4^m +5 )} &= \displaystyle{\frac{1}{9}\left((10^{m+1}-10^m) +3\cdot (4^{m+1}-4^m) \right)} \\
 &=\displaystyle{\frac{1}{9}\left[\left(10^m(10-1)+3 \cdot 4^m (4-1)  \right)\right]}\\
 &= \left(10^m+4^m\right)
\end{align*}
$$
which is an integer, and therefore ${\displaystyle{\frac{1}{9}}(10^{m+1}+3 \cdot 4^{m+1} + 5)}$ is an integer.
