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? 

When we are given an expression of the form:
$$\frac{k(k+1)(2k+1)}{6}  + (k + 1)^2$$
We should recognize that this is a case of simply adding fractions which can also be referred to as rational expressions.
The first thing I would suggest is to rewrite this as a case of adding fractions:
$$\dfrac{k(k+1)(2k+1)}{6}  + \dfrac{(k + 1)^2}{1}$$
Notice the denominators are different. In order to add fractions or rational expressions we need a common denominator. In this case, a common denominator could be 6. 
What we will do is multiply the numerator and the denominator $\dfrac{(k + 1)^2}{1}$ by 6. We should also expand $(k+1)^2=(k+1)(k+1)$.
Rewriting the equation and having a common denominator of 6:
$$\dfrac{k(k+1)(2k+1)}{6}  + \dfrac{6(k + 1)(k+1)}{1(6)}$$
Now that we have a common denominator of 6 we can simply add the fraction and simplify.
$$\dfrac{k(k+1)(2k+1)+6(k + 1)(k+1)}{6} $$
$$=\dfrac{k(2k^2+3k+1)+6(k^2+2k+1)}{6} $$
$$=\dfrac{(2k^3+3k^2+k)+(6k^2+12k+6)}{6} $$
$$=\dfrac{2k^3+9k^2+13k+6}{6} $$
Factoring again:
$$=\dfrac{2k^3+9k^2+13k+6}{6} $$
Giving us our final answer simplified:
$$\boxed{\dfrac{(k+1)(k+2)(2k+3)}{6}}$$
