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

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 $(k+1)$.

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

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

• As written, that statement is false. Plug in $k = 5$ to see that; the left side is $12$, the right side is $3$. Dec 14, 2014 at 0:03
• Can the left hand side be rewritten to resemble the right? Dec 14, 2014 at 0:04
• @DanielPareja Yes: just delete the $\;k+1\;$ on the left hand. :>) Dec 14, 2014 at 0:06
• Nope. The only values of $k$ for which the claim is true are $k = -1, -2$. I think you have to assume that it's a typo (or a "braino") on the part of the textbook author and move on to the next problem. (As a textbook author myself, I can assure you that such things happen!) Dec 14, 2014 at 0:06
• This statement is false as is. Are you sure the value of k is not somehow restricted or that you copied this correctly?
– 123
Dec 14, 2014 at 0:06

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}