Convert the following: $\frac{-1^{k}(k+1)}{2}(-k-2)$ to $\frac{-1^{k+1}(k+2)}{2}$

Question

I am teaching myself induction proofs and stepping through the algebra for the sample problems. But I got stuck on this part, can't get rid of $(k+1)$.

Can someone please step me through the process (or provide helpful links)? Thank you in advance.

$$\frac{-1^{k+1}(k+1)}{2}(k+2)$$ to

$$ \frac{-1^{k+1}(k+2)}{2}$$

EDIT sorry, my problem is to convert left formula to the right.

EDIT 2 Problem I am looking at (first page), and bellow is the how it is shown on the example.

$$\frac{-1^{k}(k+1)}{2}(-k-2)$$

to

$$\frac{-1^{k+1}(k+2)}{2}$$

Answer 1

Surely @columbus is correct in his/her inference: the actual problem should be this:

Show that $$ \frac{-1^{k}(k+1)}{2}(-k-2) $$ is the same as $$ \frac{-1^{k+1}(k+1)(k+2)}{2}. $$

To show this, look at the following sequence of steps: \begin{align} \frac{-1^{k+1}(k+1)(k+2)}{2} &=\frac{-1^{k}(-1)^1(k+1)(k+2)}{2} \text{, because $a^{b+c} = a^b a^c$}\\ &=\frac{-1^{k}(-1)(k+1)(k+2)}{2} \text{, because $-1^1 = -1$}\\ &=\frac{-1^{k}(k+1)((-1)k+(-1)2)}{2} \text{, by distributive law}\\ &=\frac{-1^{k}(k+1)(-k-2)}{2} \text{, by arithmetic}\\ &=\frac{-1^{k}(k+1)}{2}(-k-2) \text{, by commutativity of multiplication.}\\ \end{align}