# Math Help - Sigma Notation

1. ## Sigma Notation

Write the indicated sum in sigma notation:

1-(1/2)+(1/3)-(1/4)+...-(1/100)

Another one using Special Sum Formulas
nEi=1 (2i-3)^2

I wanted to break it up but the square is throwing me off

btw the n is on top of the E and i=2 is below it.

2. Originally Posted by pakman
Write the indicated sum in sigma notation:

1-(1/2)+(1/3)-(1/4)+...-(1/100)

Another one using Special Sum Formulas
nEi=1 (2i-3)^2

I wanted to break it up but the square is throwing me off

btw the n is on top of the E and i=2 is below it.
$(-1)^0\frac{1}{1}+(-1)^1\frac{1}{2}+(-1)^2\frac{1}{3}+...$
Thus,
$\sum_{k=1}^{100}(-1)^{k-1}\frac{1}{k}$

3. Thanks but it still makes no sense to me

4. Originally Posted by pakman
Thanks but it still makes no sense to me
What is your problem?

5. Originally Posted by ThePerfectHacker
What is your problem?
I don't even know where to begin with that problem. The only problems that we've gone over in class were ones that went in sequence... this one seems out of whack I guess.

6. Originally Posted by pakman
I don't even know where to begin with that problem. The only problems that we've gone over in class were ones that went in sequence... this one seems out of whack I guess.
Well, do you even know how a sigma works???

7. Originally Posted by ThePerfectHacker
Well, do you even know how a sigma works???
I think I understand the basic concept of it, however I don't seem to have a full grasp of it once it gets a little complex. I know what each variable does... however something like what you did for my problem seems confusing. I don't see how your solution solves my problem

8. Originally Posted by pakman
I think I understand the basic concept of it, however I don't seem to have a full grasp of it once it gets a little complex. I know what each variable does... however something like what you did for my problem seems confusing. I don't see how your solution solves my problem
You start from one and you have,
$(-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1$---> first term good
$(-1)^{2-1}\frac{1}{2}=(-1)^1\frac{1}{2}=-1/2$---> second term good
$(-1)^{3-1}\frac{1}{3}=(-1)^2\frac{1}{3}=1/3$---> thrid term good.
And so on

9. Originally Posted by ThePerfectHacker
You start from one and you have,
$(-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1$---> first term good
$(-1)^{2-1}\frac{1}{2}=(-1)^1\frac{1}{2}=-1/2$---> second term good
$(-1)^{3-1}\frac{1}{3}=(-1)^2\frac{1}{3}=1/3$---> thrid term good.
And so on
Thanks, that helped a lot!

10. You wouldn't by chance know how to solve the second problem using a Special Sum Formula?

11. Nevermind I figured it out. Square it out to 4i^2-12i+9 and just break it up from there

12. Hello, pakman!

Another one using Special Sum Formulas:
. . $\sum^n_{i=1} (2i-3)^2$

I'm only guessing, but are these the Special Sum Formulas?

. . $\sum^n_{i=1}1\;=\;n \qquad \sum^n_{i=1}i\;=\;\frac{n(n+1)}{2} \qquad \sum^n_{i=1}i^2\;=\;\frac{i(i+1)(2i+1)}{6}$

We have: . $\sum^n_{i=1}(2i-3)^2\;=\;\sum^n_{i=1}\left(4i^2 - 12i + 9\right)$

. . Then: . . . . $4\sum^n_{i=1}i^2 \quad\;\;- \quad\;\;12\sum^n_{i=1}i \quad+ \quad9\sum^n_{i=1}1$

. . . . . . . . . . . . $\downarrow$ . . . . . . . . . . $\downarrow$ . . . . . . . $\downarrow$

. . $= \;4\overbrace{\left[\frac{n(n+1)(2n+1)}{6}\right]} - 12\overbrace{\left[\frac{n(n+1)}{2}\right]} \;+ \;9\overbrace{[n]}$

which simplifies to: . $\frac{n}{3}\left(4n^2 - 12n + 11)$