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 $6$