Re-arranging equation.

[bubokribuck]

I need to rearrange the equation of km(m-1)+2m2 so it will look like km(m+1) in the end.

I start by expanding the brackets:

km(m-1)+2m2
= km2-km+2m2
= km(m-1)+2m2 (1)
or m(km-k+2m) (2)
or m2(2+k)-km (3)

Then I'm stuck. I've tried many other ways but they all become either (1) or (2) or (3) at the end. I have to find a way to get the final result as km(m+1), please help!

[gb7nash]

I need to rearrange the equation of km(m-1)+2m2 so it will look like km(m+1) in the end.


I hate to break it to you, but these two expressions are not algebraically equivalent.

Let k = 0, m = 1.

[ArcanaNoir]

I hate to break it to you, but these two expressions are not algebraically equivalent.

Let k = 0, m = 1.

This is true, but if you mean to say you want the result to be km(m+1) + other stuff, then I suggest you try adding and subtracting km to the expanded expression, and then gathering km(m+1) from there.

[bubokribuck]

I must have done something wrong then. Will there be any solutions if k=(-1)m-1? I set k=(-1)m-1 to make life easier, maybe this is where I went wrong.

So is there a way to arrange (-1)m-1m(m-1)+2m2 into (-1)m-1m(m+1)?

[gb7nash]

So is there a way to arrange (-1)m-1m(m-1)+2m2 into (-1)m-1m(m+1)?

Unfortunately, no. Try m = 3 for both and you'll obtain different answers for each.

[bubokribuck]

Can't figure where I've done wrong, so I thought I'd post the whole question and start by scratch.

Q: Show the sum of the first m terms of the series 12 - 22 + 32 - 42 + ... is 0.5(-1)m-1m(m+1) (for m any positive integer).

This is what I've done so far.

12 - 22 + 32 - 42 + ... + (m-1)2
= 0.5(-1)m-1-1(m-1)(m-1+1)
= 0.5(-1)m-2(m)(m-1)

Add -m2 to both sides:

12 - 22 + 32 - 42 + ... + (m-1)2 - m2 = 0.5(-1)m-2(m)(m-1)-m2

As the above equation represents the sum of the first m terms, so it must mean that
0.5(-1)m-2(m)(m-1)-m2 = 0.5(-1)m-1m(m+1) and if I can prove this then I'm done.

However, I figured that there's something wrong with the equation 0.5(-1)m-2(m)(m-1)-m2 as it's only true for when m is even. For any m that's odd, it doesn't work.

So what exactly have I done wrong?

[ArcanaNoir]

0.5(-1)m-2(m)(m-1)-m2 = 0.5(-1)m-1m(m+1)

Huh?

[bubokribuck]

Huh?

Is it not correct? I sincerely don't know where I've gone wrong. If you have spotted any errors could you please let me know?

[eumyang]

Can't figure where I've done wrong, so I thought I'd post the whole question and start by scratch.

Q: Show the sum of the first m terms of the series 12 - 22 + 32 - 42 + ... is 0.5(-1)m-1m(m+1) (for m any positive integer).

This is what I've done so far.

12 - 22 + 32 - 42 + ... + (m-1)2
= 0.5(-1)m-1-1(m-1)(m-1+1)
= 0.5(-1)m-2(m)(m-1)

Add -m2 to both sides:
Here you're making the assumption that m is even. If m was odd, you would add +m2. I haven't tried to figure this out, but maybe you could add (-1)m-1m2 instead to take care of the signs.

[eumyang]

I think I figured it out.
Can't figure where I've done wrong, so I thought I'd post the whole question and start by scratch.

Q: Show the sum of the first m terms of the series 12 - 22 + 32 - 42 + ... is 0.5(-1)m-1m(m+1) (for m any positive integer).

This is what I've done so far.

12 - 22 + 32 - 42 + ... + (-1)m-2(m-1)2
= 0.5(-1)m-1-1(m-1)(m-1+1)
= 0.5(-1)m-2(m-1)(m)
You'll need the part that's bolded above as well. You're assuming that (m-1) is odd.
Add -m2 to both sides:
As I've said, change to "add (-1)m-1m2 to both sides".
12 - 22 + 32 - 42 + ... + (-1)m-2(m-1)2 + (-1)m-1m2 = 0.5(-1)m-2(m-1)(m) + (-1)m-1m2

Now simplify the RHS to get 0.5(-1)m-1m(m+1).

[Mark44]

Can't figure where I've done wrong, so I thought I'd post the whole question and start by scratch.

Q: Show the sum of the first m terms of the series 12 - 22 + 32 - 42 + ... is 0.5(-1)m-1m(m+1) (for m any positive integer).

This is what I've done so far.

12 - 22 + 32 - 42 + ... + (m-1)2
= 0.5(-1)m-1-1(m-1)(m-1+1)
= 0.5(-1)m-2(m)(m-1)

Add -m2 to both sides:

12 - 22 + 32 - 42 + ... + (m-1)2 - m2 = 0.5(-1)m-2(m)(m-1)-m2

As the above equation represents the sum of the first m terms, so it must mean that
0.5(-1)m-2(m)(m-1)-m2 = 0.5(-1)m-1m(m+1) and if I can prove this then I'm done.

No, what you have above is the sum of the first m - 1 terms of the series. You are apparently doing a proof by induction, but you have not made any mention of this fact.



However, I figured that there's something wrong with the equation 0.5(-1)m-2(m)(m-1)-m2 as it's only true for when m is even. For any m that's odd, it doesn't work.

So what exactly have I done wrong?

[Mark44]

Here you're making the assumption that m is even. If m was odd, you would add +m2. I haven't tried to figure this out, but maybe you could add (-1)m-1m2 instead to take care of the signs.
eumyang, you have that backwards. The even terms are negative, and the odd terms are positive.

[bubokribuck]

I think I figured it out.

You'll need the part that's bolded above as well. You're assuming that (m-1) is odd.

As I've said, change to "add (-1)m-1m2 to both sides".
12 - 22 + 32 - 42 + ... + (-1)m-2(m-1)2 + (-1)m-1m2 = 0.5(-1)m-2(m-1)(m) + (-1)m-1m2

Now simplify the RHS to get 0.5(-1)m-1m(m+1).

No, what you have above is the sum of the first m - 1 terms of the series. You are apparently doing a proof by induction, but you have not made any mention of this fact.

I have managed to solve the problem using proof by induction, thanks for your help :)

[eumyang]

eumyang, you have that backwards. The even terms are negative, and the odd terms are positive.
I thought that was what I said previously. :confused:

[Mark44]

I think it actually was. I misinterpreted what you meant. I thought that you were saying referred to 12 - 22 + 32 -+ ... + m2, where m is even.