Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Induction  $(n^3+5n)$ is divisible by $6$
Here is what I have done 
Assume $n=k$ is true
$\sum_{1}^{k} k =\dfrac{(k^3+5k)}{6}$
Now assume $n=k+1$ is true
$\sum_{1}^{k+1} k+1 =\dfrac{(k+1)^3+5(k+1)}{6}$
Then now
$\dfrac{(k+1)^3+5(k+1)}{6}=\sum_{1}^{k} k + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + \dfrac{(6k+6)}{6}$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+11k+6)}{6}$
But the other side doesnt equate (LHS)
$\dfrac{(k^3+3k^2+3k+1)+5k+5}{6}$
$\dfrac{(k^3+3k^2+8k+6)}{6}\ne\dfrac{(k^3+11k+6)}{6}$
I hope my method was clear enough so you can see where I went wrong. It would be much more use to me if you solved it as I learn better from looking at solutions and then applying them to other questions. 

I don't follow the induction step. First the base case when $n=1$ is clear since $6\mid 6$. Now suppose that the claim holds when $n=k$ for some $k\geq 1$. This means that $k^3+5k=6m$ for some $m\in \mathbb{N}$. We want to show that $(k+1)^3+5(k+1)$ is divisible by $6$. Expand
$$
(k+1)^3+5(k+1)=k^3+5k+6+3(k^2+k)
$$
and use the induction hypothesis, as well as the fact that the product of two consecutive integers is even.
Alternatively, we can express the polynomial in the binomial coefficient basis using the fact that
$$
n^k=\sum_{j=0}^k S(k,j)n^{\underline{j}}
=\sum_{j=0}^k S(k,j)\binom{n}{j}j!
$$
where $S(k,j)$ is the stirling number of the second kind to yield that
$$
n^3+5n=6\binom{n}{1}+6\binom{n}{2}+6\binom{n}{3}
$$
