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. 

When you start, you assume the given statement is true for $n=k$. The given statement is that $n^3 + 5n$ is divisible by $6$. Then you have to assume $k^3+5k$ is divisible by $6$, i.e., $$6N = k^3+5k$$ for some integer $N$.
There's nothing in the statement about the sum from $1$ to $n$, so the $\sum_1^k k$ term doesn't belong.
From there, you now ask whether $(k+1)^3 + 5(k+1)$ is divisible by $6$, i.e., 
$$ 6M = (k+1)^3 + 5(k+1)$$ for some integer $M$ (with no relation to $N$, yet). To demonstrate such an integer $M$ exists we expand the right side:
$$ 6M = k^3 + 3k^2 + 8k + 6$$
$$6M = (k^3 + 5k) + (3k^2 + 3k + 6)$$
$$6M = 6N + (3k^2 + 3k + 6)$$
$$M = N + \frac{3k^2+3k+6}{6} = N + \frac{k^2+k+2}{2} = N+1+\frac{k^2+k}{2}$$
So the problem now reduces to showing that $k^2+k$ is divisible by $2$. But $k^2$ and $k$ have the same parity, so their sum is always even; thus we're done.
