<s>[INST] Is this a proper use of induction? ($(n^2+5)n$ is divisible by 6) Just want to get input on my use of induction in this problem:
Question. Use mathematical induction to prove that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$. 
Proof by mathematical induction.
(1) show base case ($n=1$) is true: 
$$
((1)^2 + 5) (1) = 6
$$
$6$ is indeed divisible by $6$, so base case ($n=1$) is true
(2a) Assume case $n$ is true: $(n^2+5)n$ is divisible by $6$.
(2b) Show that case $n$ $\implies$ case $(n+1)$
$$
\begin{align*}
((n+1)^2+5)(n+1) 
&\rightarrow 
((n^2+2n+1)+5)(n+1)  
\\ &\rightarrow  
[(n^2+5)+(2n+1)](n+1)
\\ &\rightarrow  
(n^2+5)n + (n^2+5)+(2n+1)n+ (2n+1)
\\ &\rightarrow  
(n^2+5)n + [(n^2+5)+(2n^2+n)+ (2n+1)]
\\ &\rightarrow  
(n^2+5)n + [(3n^2+3n)+6]
\end{align*} 
$$
Now we can see case $(n+1)$ $= (n^2+5)n + (3n^2+3n)+6$. 
We know $6$ is divisible by $6$ and we are assuming $(n^2+5)n$ is divisible by $6$ already, so all we need to do is show $(3n^2+3n)$ is divisible by $6$:
Letting $n=1$ for $(3n^2+3n)$ gives: $(3(1)^2+3(1)) = 6$
Thus, it has been demonstrated that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$. 
I'm not sure if letting $n=1$ for that last expression is enough to prove it is divisible by $6$
 [/INST] You can just say that it is always divisible by 6 since it is divisible by 3 (from the expression) and it is divisible by 2 since it is a multiplication of two consecutive integers- one of which is even. 
Also the whole problem could be solved very easily using divisibility rather than induction, but I guess your problem asked you to use it.
</s>