Use mathematical induction to prove an assertion The assertion: $n^3 + 5n$ is divisible by $6$
I have completed the basis step $(n=1)$ and the first part of the induction step $(n=k)$, but I am stuck on the second part $(n=k+1)$.
This is what I have so far:
For $n=k$: $k^3 + 5k = 6t$
For $n= k+1$:
$(k+1)^3 + 5(k+1)$
$= k^3 + 3k^2 + 8k + 6$
$= (k^3 + 5k) + 3k + 3k^2 + 6$
$= 3k^2 + 3k + 6t + 6$
$= ???$
I cannot pull out a $6$ because that would leave halves that cannot be counted as integers. How should I proceed?

You can complete your work this way:
$$
(k+1)^3 + 5(k+1)=(k^3 + 5k) + 3k + 3k^2 + 6=6t+3k+3k^2+6=6 \left(t+1+\frac{k(k+1)}{2} \right)
$$
and note that $k$ or $k+1$ is even.
