# Thread: Calculating Sums

1. ## Calculating Sums

Is there some sort of trick when dealing with the following sum?:
$\sum_{k=2} ^{99} (-1)^k(k)^2$

I attempted it numerous times, but the (-1)^k is messing me up

2. $\sum_{k=2} ^{99} (-1)^k(k)^2 = \sum_{k=1} ^{49} ((2k)^2 - (2k+1)^2)$
$= \sum_{k=1} ^{49} ((2k) - (2k+1))((2k) + (2k+1))=\sum_{k=1} ^{49} -(4k+1)$

3. Originally Posted by quantoembryo
Is there some sort of trick when dealing with the following sum?:
$\sum_{k=2} ^{99} (-1)^k(k)^2$

I attempted it numerous times, but the (-1)^k is messing me up
$(-1)^2\;2^2+(-1)^3\;3^2+(-1)^4\;4^2+.......$

$=2^2-3^2+4^2-5^2+....$

$=\displaystyle\sum_{k=1}^{49}(2k)^2-\sum_{k=1}^{49}(2k+1)^2$

4. So when you are tackling these sort of problems, is it always best to set k=1?

5. For this one, the theme is if you have $(-1)^k$ (and can't simplify directly) split the sum into the positive parts and the negative parts.

6. Originally Posted by quantoembryo
So when you are tackling these sort of problems, is it always best to set k=1?
If k is even

$(-1)^k=1$

If k is odd

$(-1)^k=-1$

That will simplify the situation.

7. I really don't know why I am having so much trouble with this, but I just am not seeing it. Which summation formula(s) is being used?

8. $\sum_{k} (-1)^kf(k) = \sum_{k\, even}f(k) - \sum_{k\, odd}f(k)$
Does this make sense?
Or are you confused about another part?

9. Yes, that helps substantially. Thanks a lot!

10. Originally Posted by quantoembryo
Is there some sort of trick when dealing with the following sum?:
$\sum_{k=2} ^{99} (-1)^k(k)^2$

I attempted it numerous times, but the (-1)^k is messing me up
We can start from the known formula...

$\displaystyle S^{2}_{n} = \sum_{k=1}^{n} k^{2} = \frac{n\ (n+1)\ (2n+1)}{6}$ (1)

... and from (1) derive with some 'easy' step...

$\displaystyle \sum_{k=2}^{99} (-1)^{k} k^{2} = 1-S^{2}_{99} + 8\ S^{2}_{49} =- 4949$ (2)

Kind regards

$\chi$ $\sigma$