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.
$\displaystyle (-1)^0\frac{1}{1}+(-1)^1\frac{1}{2}+(-1)^2\frac{1}{3}+...$
Thus,
$\displaystyle \sum_{k=1}^{100}(-1)^{k-1}\frac{1}{k}$

3. Thanks but it still makes no sense to me

4. Originally Posted by pakman
Thanks but it still makes no sense to me

5. Originally Posted by ThePerfectHacker
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,
$\displaystyle (-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1$---> first term good
$\displaystyle (-1)^{2-1}\frac{1}{2}=(-1)^1\frac{1}{2}=-1/2$---> second term good
$\displaystyle (-1)^{3-1}\frac{1}{3}=(-1)^2\frac{1}{3}=1/3$---> thrid term good.
And so on

9. Originally Posted by ThePerfectHacker
You start from one and you have,
$\displaystyle (-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1$---> first term good
$\displaystyle (-1)^{2-1}\frac{1}{2}=(-1)^1\frac{1}{2}=-1/2$---> second term good
$\displaystyle (-1)^{3-1}\frac{1}{3}=(-1)^2\frac{1}{3}=1/3$---> 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:
. . $\displaystyle \sum^n_{i=1} (2i-3)^2$

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

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

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

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

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

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

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