# (Series) Prove the following:

• Sep 29th 2010, 11:13 AM
Quacky
(Series) Prove the following:
Show that when n is even,

$
1^3-2^3+3^3-...-n^3 \equiv 1^3 + 2^3 + 3^3 +...+ n^3 - 16(1^3 + 2^3 + 3^3 +...+(\frac{n}{2})^3) \equiv \sum\limits_{r=1}^{n} r^3 - 16 \sum\limits_{r=1}^{\frac{n}{2}} r^3$

I don't know how to approach this question, the 'when n is even' is throwing me, and even without that part, I think I'd struggle. How should I begin?
• Sep 29th 2010, 11:35 AM
HappyJoe
Did you try proving it by induction? It seems fit for this kind of problem.
• Sep 29th 2010, 11:37 AM
Quacky
No, I didn't try induction. The reason is that this chapter is before induction in the textbook, and we didn't cover induction in class untill after the work was set. I suppose it's worth a try though.
• Sep 29th 2010, 11:51 AM
Traveller
Did you cover the formula for summing cubes ?
• Sep 29th 2010, 12:03 PM
Quacky
Yes Traveller, if by that you mean :

$\sum\limits_{r=1}^{n} r^3 = \frac{n^2}{4}(n+1)^2$

But I still can't see a way of applying it to this question specifically.
• Sep 29th 2010, 12:06 PM
Traveller
Quote:

Originally Posted by Quacky
Yes traveller, if by that you mean :

$\sum\limits_{r=1}^{n} r^3 = \frac{n^2}{4}(n+1)^2$

But I still can't see a way of applying it to this question specifically.

You can write each odd number as 2k-1 and each even number as 2k and then apply this formula. Then you can add and subtract suitable terms to bring it to the required form.
• Sep 29th 2010, 12:18 PM
Quacky
I'll try that now.
• Sep 29th 2010, 03:41 PM
Quacky
It seems I am still unable to solve this problem. Would you elaborate on your post above slightly? I've clearly misunderstood your method.
• Sep 30th 2010, 04:44 AM
HallsofIvy
$1- 2^3+ 3^3- 4^3+ \cdot\cdot\cdot+ (n-1)^3+ n^3= (1+ 2^3+ 3^3+ 4^3+ \cdot\cdot\cdot+ n^3)- 2(2^3+ 4^3+ 6^3+ n^3)$
for n even.
• Sep 30th 2010, 05:54 AM
Traveller
Quote:

Originally Posted by Quacky
It seems I am still unable to solve this problem. Would you elaborate on your post above slightly? I've clearly misunderstood your method.

Sorry, I was wrong. :D

HallsofIvy is correct. You don't need anything else to prove the formula. :D
• Sep 30th 2010, 10:11 AM
Arka
Hello Quacky,
Could you please tell me the name of the book from which you got the sum you posted above ? Thanks a lot .....
• Sep 30th 2010, 11:11 AM
Quacky
Quote:

Originally Posted by HallsofIvy
$1- 2^3+ 3^3- 4^3+ \cdot\cdot\cdot+ (n-1)^3+ n^3= (1+ 2^3+ 3^3+ 4^3+ \cdot\cdot\cdot+ n^3)- 2(2^3+ 4^3+ 6^3+ n^3)$
for n even.

Yeah, I see this now! From here, I simply have to take out a factor of $2^3$ from the last part of the second expression, which gives:

$(1+ 2^3+ 3^3+ 4^3+ \cdot\cdot\cdot+ n^3)- 2(2^3)(1^3+ 2^3+ 3^3+ (\frac{n}{2})^3)$

Which can then be summarised by the third expression which I am asked to show in the question.