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 $6$
Best Answer
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,