How is this simplified? $\frac{n(2n^2+9n+1)}{6}+(n+1)^2+2(n+1)-1$

Question

How do we simplify this: $$\frac{n(2n^2+9n+1)}{6}+(n+1)^2+2(n+1)-1$$ to this: $$\frac{(n+1)(2(n+1)^2+9(n+1)+1)}{6}$$

What were the steps taken to get to it?

Answer 1

We have

$$\frac{n(2n^2+9n+1)}{6}+(n+1)^2+2(n+1)-1$$

$$\frac16\left(2n^3+9n^2+n+6(n+1)^2+12(n+1)-6\right)$$

$$\frac16\left(2n^3+6n^2+6n+2+3n^2-5n-2+6(n+1)^2+12(n+1)-6\right)$$

$$\frac16\left(2(n+1)^3+3(n^2+2n+1)-11n-5+6(n+1)^2+12(n+1)-6\right)$$

$$\frac16\left(2(n+1)^3+9(n+1)^2+(n+1)\right)$$

$$\frac{(n+1)(2(n+1)^2+9(n+1)+1)}{6}$$

Answer 2

Work backwards and distribute:

$(\color{blue}n+1)(\color{blue}2(\color{blue}n+1)^2 + \color{blue}9(\color{blue}n+1) + 1) = $

$\color{blue}n(\color{blue}2(\color{blue}n+1)^2 + \color{blue}9(\color{blue}n+1) + 1) + (2(n+1)^2 + 9(n+1) + 1)=$

$\color{blue}{n(2n^2 + 9n +1)} + (2(n+1)^2 + 9(n+1) + 1) + 2n(2n+1) +9n$

Meanwhile

$(2(n+1)^2 + 9(n+1) + 1) + 2n(2n+1) +9n =$

$(2(n+1)^2+ 9(n+1) + 1) +4n^2 + 11n=$

$(2(n+1)^2+ 9(n+1) + 1) +4n^2 + 8n+4+3n-4 =$

$(2(n+1)^2+ 9(n+1) + 1)+ 4(n+1)^2 +3n -4=$

$(2(n+1)^2+ 9(n+1) + 1)+ 4(n+1)^2 +3(n+1) -7=$

$6(n+1)^2 + 12(n+1) - 6=$

$6[\color{green}{(n+1)^2 + 2(n+1) - 1}$

....

So $\frac {(n+1)(2(n+1)^2 + 9(n+1) + 1)}6 =$

$\frac {\color{blue}{n(2n^2 + 9 + 1)}}6 + \frac {6(\color{green}{(n+1)^2 + 2(n+1) - 1})}6$

$=\frac {\color{blue}{n(2n^2 + 9 + 1)}}6 + \color{green}{(n+1)^2 + 2(n+1) - 1}$