# summations

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• Aug 31st 2008, 08:33 PM
iplayaltosax
summations
Well, I had a summer homework packet that I put off, and now I really need to finish it fast... Anyway, I know how summation notation works, but in my packet there's a problem that says "Find Each summation" and where there's usually a number over the sigma letter, there's an infinity symbol. I never learned this last year, so I don't know what to do. Help?
• Aug 31st 2008, 08:36 PM
Chris L T521
Quote:

Originally Posted by iplayaltosax
Well, I had a summer homework packet that I put off, and now I really need to finish it fast... Anyway, I know how summation notation works, but in my packet there's a problem that says "Find Each summation" and where there's usually a number over the sigma letter, there's an infinity symbol. I never learned this last year, so I don't know what to do. Help?

It represents an infinite series. What is the exact problem? Post it here so we have a better idea of how to help you.

--Chris
• Aug 31st 2008, 08:37 PM
Jhevon
Quote:

Originally Posted by iplayaltosax
Well, I had a summer homework packet that I put off, and now I really need to finish it fast... Anyway, I know how summation notation works, but in my packet there's a problem that says "Find Each summation" and where there's usually a number over the sigma letter, there's an infinity symbol. I never learned this last year, so I don't know what to do. Help?

there are various ways to tackle problems like these based on the summation you have. it would be good for you to state the exact problem. otherwise we might go into explaining things that are not necessary to explain, re-inventing the wheel, etc
• Aug 31st 2008, 08:42 PM
iplayaltosax
Well, below the sigma letter it says i=1, then off to the right it says 3(1/2)^(i-1). That better?
• Aug 31st 2008, 08:49 PM
Chris L T521
Quote:

Originally Posted by iplayaltosax
Well, below the sigma letter it says i=1, then off to the right it says 3(1/2)^(i-1). That better?

$\displaystyle \sum_{i=1}^{\infty}3\left(\frac{1}{2}\right)^{i-1}$

This is a geometric series, where $\displaystyle a=3$ and $\displaystyle r=\frac{1}{2}$

Since $\displaystyle \left|\frac{1}{2}\right|<1$, the series will converge to a finite sum, and that sum is $\displaystyle \frac{a}{1-r}$

Try to take it from here.

--Chris
• Sep 1st 2008, 09:39 AM
iplayaltosax
Quote:

Originally Posted by Chris L T521
$\displaystyle \sum_{i=1}^{\infty}3\left(\frac{1}{2}\right)^{i-1}$

This is a geometric series, where $\displaystyle a=3$ and $\displaystyle r=\frac{1}{2}$

Since $\displaystyle \left|\frac{1}{2}\right|<1$, the series will converge to a finite sum, and that sum is $\displaystyle \frac{a}{1-r}$

Try to take it from here.

--Chris

Thanks, that helped on that one. I get it, but the next problem says

$\displaystyle \sum_{i=1}^{n}(i^2 -3i+2)$
• Sep 1st 2008, 10:29 AM
Jhevon
Quote:

Originally Posted by iplayaltosax
Thanks, that helped on that one. I get it, but the next problem says

$\displaystyle \sum_{i=1}^{n}(i^2 -3i+2)$

recall that the summation of a sum is just the sum of the summations. and also that we can factor constants out and put them in front of the summation symbol. thus,

$\displaystyle \sum_{i = 1}^n (i^2 - 3i + 2) = \sum_{i = 1}^ni^2 -3 \sum_{i = 1}^ni + \sum_{i = 1}^n2$

and recall the rules:

$\displaystyle \sum_{i = 1}^ni^2 = \frac {n(n + 1)(2n + 1)}6$, $\displaystyle \sum_{i = 1}^ni = \frac {n(n + 1)}2$, and $\displaystyle \sum_{i = 1}^nc = cn$ for $\displaystyle c$ a constant

i think you can take it from here
• Sep 1st 2008, 10:41 AM
Moo
Hello !
Quote:

Originally Posted by iplayaltosax
Thanks, that helped on that one. I get it, but the next problem says

$\displaystyle \sum_{i=1}^{n}(i^2 -3i+2)$

Hmmm I have 2 possibilities for this... The second one being unsure, it's the first time (Worried)

1st :
Note that $\displaystyle \sum_{i=1}^n (i^2-3i+2)=\sum_{i=1}^n i^2-3 \sum_{i=1}^n i+2 \sum_{i=1}^n 1$

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

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

-------------------------------------------
2nd : you gotta be used to summations, so don't read if you don't want to... The first part is largely enough
$\displaystyle i^2-3i+2=(i-2)(i-1)$

$\displaystyle S=\sum_{i=1}^n (i^2-3i+2)=\sum_{i=1}^n (i-2)(i-1)$

We'll try to write this sum in the form $\displaystyle \sum_{i=\dots}^{\dots} (i-1)(i)$
Change i into t+1.
i-2 --> t-1
i-1 --> t
i=1 --> t=0
i=n --> t=(n-1)
Then $\displaystyle S=\sum_{i=1}^n (i-2)(i-1)=\sum_{t=0}^{n-1} (t-1) \cdot t$

Thus \displaystyle \begin{aligned} 2S &=\sum_{i=1}^n (i-2)(i-1)+\sum_{t=0}^{n-1} (t-1) \cdot t \\ &=\sum_{i=1}^n (i-2)(i-1)+\sum_{i=0}^{n-1} (i-1) \cdot i \\ &=\left(\sum_{i=1}^{{\color{red}n-1}} (i-2)(i-1)\right)+{\color{red}(n-2)(n-1)}+\left(\sum_{{\color{red}i=1}}^{n-1} (i-1) \cdot i\right)+{\color{red}0} \end{aligned}

$\displaystyle 2S=(n-2)(n-1)+\sum_{i=1}^{n-1} \bigg[(i-2)(i-1)+(i-1)i\bigg]$

$\displaystyle 2S=(n-2)(n-1)+\sum_{i=1}^{n-1} (i-1)\underbrace{(i-2+i)}_{2(i-1)}$

$\displaystyle 2S=(n-2)(n-1)+2 \sum_{i=1}^{n-1} (i-1)^2$

Changing i again :

$\displaystyle 2S=(n-2)(n-1)+2 \sum_{t=0}^{n-2} t^2$

$\displaystyle 2S=(n-2)(n-1)+2 \left(\frac{(n-2)(n-1)(2n-3)}{6}\right)$ By applying the formula above.

$\displaystyle 2S=(n-2)(n-1) \left(1+\frac{2n-3}{3}\right)$

And so on... it's much longer than the normal method (Rofl)

Edit : no wonder I'm too late -___-' what an idea to make such a useless stuff lol !
• Sep 1st 2008, 10:51 AM
iplayaltosax
Quote:

Originally Posted by Jhevon
recall that the summation of a sum is just the sum of the summations. and also that we can factor constants out and put them in front of the summation symbol. thus,

$\displaystyle \sum_{i = 1}^n (i^2 - 3i + 2) = \sum_{i = 1}^ni^2 -3 \sum_{i = 1}^ni + \sum_{i = 1}^n2$

and recall the rules:

$\displaystyle \sum_{i = 1}^ni^2 = \frac {n(n + 1)(2n + 1)}6$, $\displaystyle \sum_{i = 1}^ni = \frac {n(n + 1)}2$, and $\displaystyle \sum_{i = 1}^nc = cn$ for $\displaystyle c$ a constant

i think you can take it from here

Wow. Thanks a lot. I'm positive that I never learned this in the school year, so I was wondering where I could find these rules and maybe some explanations... Would you know where?
• Sep 1st 2008, 11:02 AM
Jhevon
Quote:

Originally Posted by iplayaltosax
Wow. Thanks a lot. I'm positive that I never learned this in the school year, so I was wondering where I could find these rules and maybe some explanations... Would you know where?

well, if you want to find out about summation rules, googling "summation rules" is always a good idea. :p here is a page i found by doing that: http://www.math.ucdavis.edu/~kouba/C...Summation.html

otherwise you can check well known sources like wikipedia.org, and maybe it is on purplemath