Rearranging Equation: Solving for km(m+1) from km(m-1)+2m^2

[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.

 

[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.