<s>[INST] Prove $3^{2n+1} + 2^{n+2} = 7m$ I'm trying to prove that $3^{2n+1} + 2^{n+2}$ is a multiple of 7 by using induction. 
So I started to prove it for $n=1$: $3^{2(1)+1}+2^{1+2}=3^3+2^3=27+8=35=7(5)$. 
Next, try to prove that the statement being true $n=k$ implies it being true for $n=k+1$. Thus: 
$3^{2(k+1)+1}+2^{(k+1)+2} = 3^{2k+3}+2^{k+3} = (3)(3)(3^{2k+1})+(2)(2^{k+2}) = 9(3^{2k+1})+2(2^{k+2})$
I feel like I'm almost there, if I could've factor 9 and 2 somehow I could say that $3^{2k+1}+2^{k+2}=7m$ for some integer $m$, but I can't find a way to do it. What am I missing? Or did I do a blunder somewhere along the road?
Thanks in advance.
 [/INST] the trick to induction is put the $k+1$ expression in terms of the $k$ expression.
If $ 3^{2k+1} + 2^{k+2}=7m$ then
$3^{2(k+1)+1}+2^{(k+1)+2} =  9(3^{2k+1})+2(2^{k+2}) = 2[ 3^{2k+1} + 2^{k+2}] + 7*3^{2k+1}$
$= 2(7m) + 7*3^{2k+1} = 7(2m + 3^{2k+1})$.
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