Question

PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!
I WILL AWARD MEDAL!!!!!!!!!!!!!!!!!!!!!!!!!!!

Can some one explain to me what is sigma notation.

Answer 1

\[\sum_{}^{} \]

Answer 2

Sigma is a sum. Sigma notation asks you to some something.
\[\sum_{i=1}^{5} x\] means sum all numbers i from 1 to 5. 1+2+3+4+5 = 15.

Answer 3

The \[\sum_{}^{}\] is sigma notation.

Answer 4

It means the sum of, ie the addition of all terms.

Answer 5

Ok and what does the i=1 stand for? Does that refer to imaginary numbers or what does it refer to?

Answer 6

It has absolutely nothing to do with imaginary numbers. It just means for i = 1 to i = 5, sum all i's. Like what I said above. (The x is supposed to be an i.)

Answer 7

Ok can you solve one example with showing the step please so I can understand.

Answer 8

Given the arithmetic sequence –6, 1, 8…, evaluate \[\sum_{n=9}^{12}a _{n}\]

Answer 9

\[\Large \sum_{k=1}^{4} x^2 \]

this means the sum of all x^2's, starting from 1 and ending at 4. It doesn't matter if you use k, or i or whatever letter. \[\Large \sum_{k=1}^{4} x^2= 1^2 + 2^2 +3^2 +4^2 = \]

Answer 10

@Loser66

Answer 11

\[\Large \sum_{n=9}^{12}a _{n} = a_9 + a_{10} +a_{11} + a_{12}\]
So you need to first find a9, a10, a11, a12.

What's the common difference (how much is added from one term to the next) of: –6, 1, 8

Answer 12

You add 7

Answer 13

So what is next.

Answer 14

Can you find the general rule? a1 is the first term, d is the difference you just found, n means the n-th term (eg if you want the 9th term, use n = 9). Use this formula to find a9 to a12, then add them up as i showed above.

\[\Large a_n = a_1 +(n-1) d\]

Answer 15

What formula.

Answer 16

Oh ok yes I see it now.

Answer 17

But what is a.

Answer 18

an means the nth term. a9 means the 9th term, so n=9. Find a9 up to a12.

Answer 19

But how if I do not know what A is.

Answer 20

an just means the nth term, whatever term (a number) of the sequence. a is not a variable you need to worry about. You have a1 (the first term) = -6, d=7, so:

\[\Large a_n = a_1 +(n-1) d\] becomes \[\Large a_n = -6 +(n-1)*7\]

Here is how you find a9 (plug in n = 9:
\[\Large a_9 = 6 +(9-1) *7 =???\]

Answer 21

Ok I have another question also if their is a - after the \[\sum -\] does that mean you are subtracting instead of adding.

Answer 22

@agent0smith can you please answer my last question I asked please.

Answer 23

Basically, yes, you'd put a negative in front of all the terms (note that a term which is already negative will become a positive).

Answer 24

Ok can we start of baisic like
\[\sum_{n=2}^{8}-(-n-5)\]

Answer 25

All you have to do is plug in each n, one by one, and add or subtract all the terms:

\[\large \sum_{n=2}^{8}-(-n-5) = -(-2-5) -(-3-5) - (-4-5) -...-(-8-5)\]

Or to make it easier, you can bring the negative outside the sum:
\[\large \sum_{n=2}^{8}-(-n-5) = - \large \sum_{n=2}^{8}(-n-5) \]
or just distribute the negative into the brackets \[\large \sum_{n=2}^{8} (n+5) = (2+5) + (3+5) + ... + (8+5)\]

Answer 26

Wait but how did n become 3.

Answer 27

By plugging in n=2, n=3, n=4... and so on.

Answer 28

That sum means start from n=2, and stop at n=8. All you have to do is plug in each n.

Answer 29

Oh so you just skip 1.

Answer 30

So the answer is 70

Answer 31

You start at whatever the bottom number is, in this case the n=2.

Answer 32

Yes I mean after you add everthing up and the final answer is 70.
can you just check my answer please.

Answer 33

http://www.wolframalpha.com/input/?i=sum+from+n%3D2+to+n%3D8+of+-%28-n-5%29++%3D

Answer 34

Yes I am correct.

Answer 35

Thank you so much.

Answer 36

You're welcome.

Answer 37

@agent0smith

Answer 38

@agent0smith

Answer 39

yes...?

Answer 40

Oh I was trying to find the answer to the first question that I found hard.

Answer 41

But I got it thank you.