Algebra simplification in mathematical induction . I was proving some mathematical induction problems and came through an algebra expression that shows as follows:
$$\frac{k(k+1)(2k+1)}{6}  + (k + 1)^2$$
The final answer is supposed to be: 
$$\frac{(k+1)(k+2)(2k+3)}{6}$$
I walked through every possible expansion; I combine like terms, simplify, factor, but never arrived at the answer.
Could someone explain the steps? 

First, let's write the expression as a sum of fractions with a common denominator.
$$\dfrac{k(k+1)(2k+1)}{6}  + (k + 1)^2 = \dfrac{k(k+1)(2k+1)}{6} + \dfrac{6(k+1)^2}{6}\tag{1}$$
Now expand $6(k+1)^2 = 6k^2 + 12k + 6\tag{2}$ and expand 
$k(k+1)(2k+1) = k(2k^2 + 3k + 1) = 2k^3 + 3k^2 + k\tag{3}$
So now, $(1)$ becomes $$\dfrac{k(k+1)(2k+1)}{6} + \dfrac{6(k+1)^2}{6} = \dfrac{(2k^3 + 3k^2 + k) + (6k^2 +12 k + 6)}{6} $$ $$= \dfrac{\color{blue}{\bf 2k^3 + 9k^2 +13k + 6}}{6}\tag{4}$$
We can factor the numerator in $(4)$, or we can expand the numerator of our "goal"...
$$\frac{(k+1)(k+2)(2k+3)}{6} = \dfrac{\color{blue}{\bf 2k^3 + 9k^2 + 13k + 6}}{6}\tag{goal}$$
