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. 

Here is a simple "direct proof":
\begin{align*}
10^n+3 \times 4^n + 5&=10^n-1 +3 \times 2^{2n}+6 =9999..9+6 \times [2^{2n-1}+1] \\
&=9999..9+6 \times (2+1)(2^{2n-2}-2^{2n-3}+\cdots-2+1) \\
&= 9 \times [1111...1+2 \times (2^{2n-2}-2^{2n-3}+\cdots-2+1)]
\end{align*}
