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
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.
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!!
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 ?
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