Algebra simplification in mathematical induction .

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"...

$$\frac{(k+1)(k+2)(2k+3)}{6} = \dfrac{\color{blue}{\bf 2k^3 + 9k^2 + 13k + 6}}{6}\tag{goal}$$

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.

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: $$\boxed{\dfrac{(k+1)(k+2)(2k+3)}{6}}$$