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. 

Hint $\rm\ \ \forall\, n\ge 0\!:\ 9\ |\ f(n)\!\iff\! [\,9\ |\ f(0)\:$ and $\rm\:\forall\, n\ge 0\!:\ 9\ |\ f(n\!+\!1)\!-\!f(n)\,]\ \ $ (typical telescopy)
Now $\rm\, f(n) = 10^n\! +\! 3\cdot 4^n\! +\! 5\ \Rightarrow\ f(0)=9,\,$ and $\rm \ 9\:|\:f(n\!+\!1)\!-\!f(n) = (10\!-\!1)\, 10^n\! +\! 3\,(4\!-\!1)\, 4^n$
Note $\,\ $ The telescopy is trivial when viewed mod $9\!:\,$ it simply says that a function on $\mathbb N$ is constant if it never changes value 
$\rm\:\forall\,n\!:\,f(n\!+\!1)\equiv f(n)\Rightarrow\forall\,n\!:\,f(n)\equiv f(0)\:$  (trivially proved by induction)
