Help with a proof that involves simple algebra

[iamsmooth]

Homework Statement

I have worked on a proof and reduced it to the following equation:

$$(-1)^k(1-4k)+2((-1)^k(1+4k))$$

To complete this part of my proof, I need to prove that algebraically, the above equation is the same as:

$$(-1)^{k+1}(1-4(k+1))$$

Homework Equations

The Attempt at a Solution

I have tried multiplying them out but the algebra doesn't seem to work (that is, I'm probably not doing it right).

Can someone just tell me if they are or AREN'T equal? I don't want to spend any more time trying to prove something that's not true.

From what I can tell, it shouldn't be equal, but I'm very unsure. You don't even have to solve it for me, just tell me if you think they're equal or not, and I'll keep trying if they are equal.

Thanks!

 

[Mark44]

Just evaluate the two expressions for several values of k.

The first one can be written as
(-1)^k[(1 - 4k) + 2(1 + 4k)] = (-1)^k(3 + 4k). Try simplifying the second expression, and think about what that extra factor of (-1) does.

 

[iamsmooth]

Wow, I didn't even think of subbing in a number and checking. Nor did I see that I could have factored out that $(-1)^k$, I knew that it was meant for changing the signs, but yeah...I'm so stupid this semester...

Anyways, I can't really simplify the second equation because that's part of what I'm trying to prove.

So I have to derive the second part from the first equation.

This is what I have now:

$$(-1)^k(1-4k)+2((-1)^k(1+4k))$$

$$=(-1)^k(1-4k)+2+8k]$$

$$=(-1)^k(3+4k)$$

$$=(-1)^k4(k+1)-1$$

$$=(-1)^{k+1}1-4(k+1)$$

I'm unsure of the last step, because I know it's the same thing, but algebraically, can I really do this? Is there a step to show how you get to the last conclusion? I tried playing around with exponent properties so that if there was an extra -1, it would be $-1^1$ and the product of that and $-1^k$ would be $-1^{k+1}$, but I don't know how to show this last step...

 

[iamsmooth]

Oh wait I think I have it:
$$=(-1)^k4(k+1)-1$$

$$=(-1)^k-1+4(k+1)$$

Now, if I factor out negative 1:

$$=(-1)^k-1[1-4(k+1)]$$

$$= (-1)^k-1^1[1-4(k+1)]$$

$$=(-1)^{k+1}(1-4(k+1))$$

Does this make sense?

 

[Mark44]

You're leaving out a lot of parentheses and/or brackets that are necessary.

Wow, I didn't even think of subbing in a number and checking. Nor did I see that I could have factored out that $(-1)^k$, I knew that it was meant for changing the signs, but yeah...I'm so stupid this semester...

Anyways, I can't really simplify the second equation because that's part of what I'm trying to prove.

So I have to derive the second part from the first equation.

This is what I have now:

$$(-1)^k(1-4k)+2((-1)^k(1+4k))$$

$$=(-1)^k(1-4k)+2+8k]$$

Above, you're missing a left bracket right after (-1)^k.

$$=(-1)^k(3+4k)$$

Above is fine.

$$=(-1)^k4(k+1)-1$$

Above, now you're missing parentheses around 4(k + 1) - 1. That whole expression is multiplied by (-1)^k.

$$=(-1)^{k+1}1-4(k+1)$$

Above, similar comment.

I'm unsure of the last step, because I know it's the same thing, but algebraically, can I really do this? Is there a step to show how you get to the last conclusion? I tried playing around with exponent properties so that if there was an extra -1, it would be $-1^1$ and the product of that and $-1^k$ would be $-1^{k+1}$, but I don't know how to show this last step...

A simpler way. What you're trying to get to is equal to (-1)^{k + 1}(-3 - 4k).

You started with:
(-1)^k(1 - 4k) + 2(-1)^k(1 + 4k)
= (-1)^k(1 - 4k + 2(1 + 4k))
= (-1)^k(3 + 4k)
= (-1)^{k + 1}(-3 - 4k) ; In this step, I multiplied by 1 in the form of (-1)(-1).
= (-1)^{k + 1}(1 - 4k - 4)
= (-1)^{k + 1}(1 - 4(k + 1))