# Thread: (Series) Prove the following:

1. ## (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?

2. Did you try proving it by induction? It seems fit for this kind of problem.

3. 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.

4. Did you cover the formula for summing cubes ?

5. 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.

6. 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.

7. I'll try that now.

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

9. $1- 2^3+ 3^3- 4^3+ \cdot\cdot\cdot+ (n-1)^3+ n^3= (1+ 2^3+ 3^3+ 4^3+ \cdot\cdot\cdot+ n^3)- 2(2^3+ 4^3+ 6^3+ n^3)$
for n even.

10. 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.

HallsofIvy is correct. You don't need anything else to prove the formula.

11. Hello Quacky,
Could you please tell me the name of the book from which you got the sum you posted above ? Thanks a lot .....

12. 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.