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.

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- Jun 22nd 2006, 02:08 PMpakmanSigma 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. - Jun 22nd 2006, 02:14 PMThePerfectHackerQuote:

Originally Posted by**pakman**

Thus,

$\displaystyle \sum_{k=1}^{100}(-1)^{k-1}\frac{1}{k}$ - Jun 22nd 2006, 02:17 PMpakman
Thanks but it still makes no sense to me :(

- Jun 22nd 2006, 02:19 PMThePerfectHackerQuote:

Originally Posted by**pakman**

- Jun 22nd 2006, 02:22 PMpakmanQuote:

Originally Posted by**ThePerfectHacker**

- Jun 22nd 2006, 02:34 PMThePerfectHackerQuote:

Originally Posted by**pakman**

- Jun 22nd 2006, 02:39 PMpakmanQuote:

Originally Posted by**ThePerfectHacker**

- Jun 22nd 2006, 02:50 PMThePerfectHackerQuote:

Originally Posted by**pakman**

$\displaystyle (-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1$---> 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 - Jun 22nd 2006, 03:08 PMpakmanQuote:

Originally Posted by**ThePerfectHacker**

- Jun 22nd 2006, 03:21 PMpakman
You wouldn't by chance know how to solve the second problem using a Special Sum Formula?

- Jun 22nd 2006, 04:13 PMpakman
Nevermind I figured it out. Square it out to 4i^2-12i+9 and just break it up from there :D

- Jun 22nd 2006, 07:51 PMSoroban
Hello, pakman!

Quote:

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