Sigma Notation

• June 22nd 2006, 02:08 PM
pakman
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.
• June 22nd 2006, 02:14 PM
ThePerfectHacker
Quote:

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,
$\sum_{k=1}^{100}(-1)^{k-1}\frac{1}{k}$
• June 22nd 2006, 02:17 PM
pakman
Thanks but it still makes no sense to me :(
• June 22nd 2006, 02:19 PM
ThePerfectHacker
Quote:

Originally Posted by pakman
Thanks but it still makes no sense to me :(

• June 22nd 2006, 02:22 PM
pakman
Quote:

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.
• June 22nd 2006, 02:34 PM
ThePerfectHacker
Quote:

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???
• June 22nd 2006, 02:39 PM
pakman
Quote:

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 :confused:
• June 22nd 2006, 02:50 PM
ThePerfectHacker
Quote:

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 :confused:

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
• June 22nd 2006, 03:08 PM
pakman
Quote:

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! :D
• June 22nd 2006, 03:21 PM
pakman
You wouldn't by chance know how to solve the second problem using a Special Sum Formula?
• June 22nd 2006, 04:13 PM
pakman
Nevermind I figured it out. Square it out to 4i^2-12i+9 and just break it up from there :D
• June 22nd 2006, 07:51 PM
Soroban
Hello, pakman!

Quote:

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