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.




  • 2
    $\begingroup$ As written, that statement is false. Plug in $k = 5$ to see that; the left side is $12$, the right side is $3$. $\endgroup$ Dec 14, 2014 at 0:03
  • $\begingroup$ Can the left hand side be rewritten to resemble the right? $\endgroup$ Dec 14, 2014 at 0:04
  • $\begingroup$ @DanielPareja Yes: just delete the $\;k+1\;$ on the left hand. :>) $\endgroup$
    – Timbuc
    Dec 14, 2014 at 0:06
  • $\begingroup$ 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!) $\endgroup$ Dec 14, 2014 at 0:06
  • $\begingroup$ This statement is false as is. Are you sure the value of k is not somehow restricted or that you copied this correctly? $\endgroup$
    – 123
    Dec 14, 2014 at 0:06

1 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}


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