Prove n-1*n = ((n-1)*n*(n+1))/3 induction Alegebra confusion

Question

Currently following a tutorial Hypothesis is k-1(k) = (k-1(k)k+1)/3

 So the induction step is 
 k-2(k-1) + k-1(k) = (k-2)(k-1)(k)/3 + k-1(k)
              = (k-2)(k-1)(k)/3 + 3(k-1)(k)/ 3
              Next step I don't understand how they got this???
              They might have skipped a few steps how did they this
              = (k-2-3)(k-1)(k)/3
              = k-1(k)(k+1) / 3 

Answer 1

In that step, they factored out $(k-1)k$ out of $$\frac{(k-2)(k-1)k+3(k-1)k}{3}$$ and you get$$\frac{k(k-1)((k-2)+3)}{3}=\frac{k(k-1)(k+1)}{3}$$as you know the point of induction is that if this formula is named $P(n)$, then you assume that it works for some arbitrary $n$ and you want to prove that $P(n) \implies P(n+1)$, you can already precalculate $P(n+1)$ to see what you need to get in the end. So in this case you should have plugged in $n+1$ and tried to do some algebraic manipulation to make it look like that. In this case you add the next term to both sides, then factor out the $k(k-1)$.