<s>[INST] 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. 
 [/INST] This is the same as proving $(10^n +3⋅4^n +5)$  is divisible by 9. The rule for divisibility by $9$ is if it's digital sum is $9$, then it's divisible by $9$  
The sum of the three parts should give a digital sum of $9$.  
$10^n$ digital sum $=1$  
$4^n$ digital sum repeats in the cycle $(4,7,1)$
multiplying these by 3 gives the cycle$(12,21,3)$
The digital sum of $12,21,3$ are each $3$
$3\times4^n$ digital sum $=3$  
$5$ is last part  
$1 +3 +5 =9$ and is therefore divisible by $9$ 
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