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"...
$$\frac{(k+1)(k+2)(2k+3)}{6} = \dfrac{\color{blue}{\bf 2k^3 + 9k^2 + 13k + 6}}{6}\tag{goal}$$
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: $$\boxed{\dfrac{(k+1)(k+2)(2k+3)}{6}}$$
Well you can make life a lot easier for yourself by using the common factor of $k+1$. Taking this out, and also the factor of a sixth for neatness, your expression is $$\frac 1 6 (k+1)(k(2k+1)+6(k+1))$$ Now you only have a quadratic to factor! $2k^2+7k+6$. You can use familiar methods to factor this as you seek.
A good idea for this sort of thing is to use Wolfram Alpha to ensure that the two things are, indeed, equal. In this case, they are, so we can spend some time looking to factor.
$$\begin{align} \frac{k(k+1)(2k+1)}{6} + (k + 1)^2 &= \frac{k(k+1)(2k+1) + 6(k + 1)^2}{6}\\ &= \frac{(k+1)(k(2k+1) + 6(k + 1))}{6}\\ &= \frac{(k+1)(2k^2 +7k + 6)}{6}\\ &= \frac{(k+1)(k+2)(2k + 3)}{6}\\ \end{align}$$
EDIT: A helpful trick is the one I did in the step from line 1 to line 2: I factored out the $k+1$ immediately, rather than expanding the whole expression. This makes it easier to deal with--most people have more practice factoring quadratics than cubics.
EDIT in response to comment: $$\begin{align} \frac{k(k+1)(2k+1) + 6(k + 1)^2}{6} &= \frac{k\color{red}{(k+1)}(2k+1) + 6\color{red}{(k+1)}(k+1)}{6}\\ &=\frac{\color{red}{(k+1)}\Big(k(2k+1) + 6(k+1)\Big)}{6} \end{align}$$
