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 $6$
 A: It is definitively not sufficient, since "letting $n=1$" means "you don't know what to do in every other cases of values of $n$". Since saying that $6$ divides $3n^2 + 3n$ is equivalent to saying that $2$ divides $n^2 + n$, you only need to show the latter. Now why should $2$ divide $n^2+n = n(n+1)$, two consecutive integers? (You could also prove this by induction, but that would be a little useless, would it.)
Hope that helps,
A: The statement that 

$3n^2+3n$ is divisible by $6$ for all $n$.

and the statement that

$3(1)^2+3(1)=6$ is divisible by $6$.

are clearly not the same :) You can prove the former statement by showing that $n^2+n=n(n+1)$ is always even (use that either $n$ is even, or $n+1$ is).
By the way, use equals signs ($=$) instead of arrows ($\rightarrow$).
Finally: for any number $k$, "case $k$" refers to the statement

$(k^2+5)k$ is divisible by $6$.

As a statement, it is true, or possibly false. Your goal is to prove that "case $n$" is true for every number $n$. Therefore, you should not think things like

"case $(n+1)$ = $(n^2+5)n+3n^2+3n+6$"

which doesn't really make sense.
A: 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.
A: $(n^{2} + 5)n \equiv (n^{2}-1)n \equiv (n-1)n(n+1) \pmod 6$. Since these are three consecutive integers, one of them must be congruent to $0\pmod 3$ and one must also be even, or congruent to $0 \pmod 2$. Then as $\gcd (2,3) = 1$, the product $(n-1)n(n+1) \equiv 0 \mod 6$. Altogether meaning that $6 | (n^{2}+5)n$ for all $n$.
