Proof explanation of $\sum_{k=0}^{n-1}(n+k)(n-k) = \frac{1}{6}n(n+1)(4n-1), n \geq 1$ I don't understand the proof of 
$$\sum_{k=0}^{n-1}(n+k)(n-k) = \frac{1}{6}n(n+1)(4n-1), n \geq 1$$
via induction.
$$\text{Base case}:  n=1 \Rightarrow \sum_{k=0}^0(1+k)(1-k)=1=\frac{1}{6}(1+1)(4-1) = \frac{6}{6}$$
$$\text{Inductive step:} \sum_{k=0}^n (n+1+k)(n+1-k) = \sum_{k=0}^n[(n+k)(n-k)+(n-k)+(n+k)+1]$$
$$= \sum_{k=0}^n (n+k)(n-k)+\sum_{k=0}^n1+2n \cdot \sum_{k=0}^n1$$
$$=\sum_{k=0}^{n-1}(n+k)(n-k)+(n+n)(n-n)+(n+1)+2n\cdot(n+1)$$
$$= \text{(IH)} \frac{1}{6}n(n+1)(4n-1)+(n+1)+2n(n+1)$$
$$ = \frac{1}{6} (n+1)(n(4n-1)+6+12n)$$
$$= \frac{1}{6}(n+1) (4n^2+11n+6)$$
$$= \frac{1}{6}(4n+3)(n+2)$$
I don't understand the first 3 lines of the inductive step.
Why do we get 
$$\sum_{k=0}^n[(n+k)(n-k)+(n-k)+(n+k)+1]$$
and 
$$= \sum_{k=0}^n (n+k)(n-k)+\sum_{k=0}^n1+2n \cdot \sum_{k=0}^n1$$
$$=\sum_{k=0}^{n-1}(n+k)(n-k)+(n+n)(n-n)+(n+1)+2n\cdot(n+1)$$

Here
$$\sum_{k=0}^n \left[(n+1+k)(n+1-k)\right] = \sum_{k=0}^n\left[(n+k)(n-k)+(n-k)+(n+k)+1\right]$$
we are using that
$$(n+1+k)(n+1-k)=((n+k)+1)((n-k)-1)=$$
$$=(n+k)(n-k)+(n-k)+(n+k)+1$$
and then since $(n+k)+(n-k)=2n$
$$\sum_{k=0}^n\left[(n+k)(n-k)+(n-k)+(n+k)+1\right]=$$
$$=\sum_{k=0}^n \left[(n+k)(n-k)\right]+$$
$$+\sum_{k=0}^n 2n+$$
$$+\sum_{k=0}^n 1$$
and for the last step, we have used that
$$\sum_{k=0}^n \left[(n+k)(n-k)\right]=\sum_{k=0}^{n-1}\left[(n+k)(n-k)\right]+(n+n)(n-n)$$
