How would I go about finding sigma notation for any given set of numbers? I really struggle at finding this. Like is there an easy way to write sigma notation just from looking at a set of numbers? Thanks

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- Nov 12th 2008, 07:17 AMBCHurricane89Sigma Notation from given set of numbers
How would I go about finding sigma notation for any given set of numbers? I really struggle at finding this. Like is there an easy way to write sigma notation just from looking at a set of numbers? Thanks

- Nov 12th 2008, 08:40 AMPlato
- Nov 12th 2008, 08:44 AMBCHurricane89
For example, your given the set of numbers: -2+4-6+8-10

Write the sum in sigma notation.

Id like to know how to figure these type of problems out, as I don't quite understand where to start. - Nov 12th 2008, 08:59 AMPlato
Examine the

**pattern**.

Every term is even and the signs alternate.

$\displaystyle \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k } \left( {2k} \right)} $

Using $\displaystyle {\left( { - 1} \right)^{k } }$ is a standard way to alternate signs.

There is no easy answer to that question. Experience and lots of practice is the key.

**Pattern recognition**is the most important skill you can have to help you. - Nov 12th 2008, 09:00 AMChris L T521
There are 5 elements, so our sum will have limits from k=1 to 5.

Next step is to try to find a pattern.

We see that the signs alternate, suggesting that there is a $\displaystyle (-1)^\alpha$ term in this sequence. Since the first term is negative, and we're starting from k=1 (which is an odd number), we see that the sequence will include a $\displaystyle (-1)^k$.

Now, what is the kth term of the sequence $\displaystyle 2,~4,~6,~8,~10$?

We note that this is the sequence of even numbers. and we know that $\displaystyle 2(1)=2,~2(2)=4,~2(3)=6,~2(4)=8,~2(5)=10$; this suggests that the kth term is $\displaystyle 2k$.

We can now see that this sum can be expressed as follows:

$\displaystyle \sum_{k=1}^5 (-1)^k (2k)=2\sum_{k=1}^5 (-1)^k k$

Does this make sense?

--Chris

EDIT: Plato beat me. As he mentioned,**pattern recognition**is very important!! - Nov 12th 2008, 09:12 AMBCHurricane89
Ok, i kinda see where your going. Whats the difference, because one of you has $\displaystyle (-1)^k$, and the other has (-1)^(k+1) is there any difference? I have a test today, and this will be on it, and I was looking at my notes, trying to figure out a way to do these, but this helps.

Oh yeah, also say teh numbers were switched, like 2-4+6-8+10, then it would still be k=1 to 5, then sigma $\displaystyle (-1)^k+1$ *2k ? - Nov 12th 2008, 09:19 AMChris L T521
There is a difference if you're starting at k=1.

If we start at k=1, we see that $\displaystyle (-1)^{k+1}$ will produce the sequence $\displaystyle 1,~-1,~1,~-1,~1,~-1,\dots$ and that $\displaystyle (-1)^k$ will produce the sequence $\displaystyle -1,~1,~-1,~1~-1,~1,\dots$

--Chris - Nov 12th 2008, 09:22 AMBCHurricane89
oh, ok, now I see that