Question

=

Answer 1

You need a sum that will involve n going from n=4 to n=9.

Also, notice that the terms where n is even are added, where n is odd they are subtracted. You can express this with a coefficient of:
\(\Large (-1)^n\)

Can you combine those hints to come up with the summation notation?

Answer 2

\[\Large \sum_{n=4}^{9}???\]

Answer 3

so its n=5?

Answer 4

No... in summation notation, you let n "run" from 1 value to another (or k, or i, or whatever "index" you want to use for the sum).

Let me do one that's similar:

\[\large (-1)^2\cdot 2^2+(-1)^3\cdot 2^3+(-1)^4\cdot 2^4+(-1)^5\cdot 2^5\\
\large \sum_{n=2}^{5}(-1)^n\cdot 2^n\\\]

Answer 5

E.g., I took expression:

4-8+16-32

....and wrote it in summation notation.

Answer 6

so (-1)^n* 2^n?

Answer 7

Now look at your expression: ¼- 1/5+ 1/6- 1/7 +1/8 – 1/9

And write it as a sum, letting n run from 4 to 9:

\[\Large \sum_{n=4}^{9}???\]

Answer 8

Hmmmm.... not quite.... the (-1)^n is part of it.

but 2^n would be:
\(\large 2^2 - 2^3 + 2^4- 2^5 + 2^6- 2^7 + 2^8- 2^9\)

That wont get you the terms of your sum.

if n=4, then 1/4 = 1/n
if n=5, then 1/5 = 1/n
etc.... see the pattern?

Answer 9

So if n is 1, then (-1)^1 gives 1, then divide by 4? so (-1)^1/n? Im confused :/

Answer 10

What's wrong with sum n = 4 9 of (1/n)*(-1)^n ?

Answer 11

Yes, I see that. First of all, (-1)^1 does not INVOLVE n. and remember, you aren't going to have n=1, at all.... your sum should run from n=4 to n=9.

Why? Because those are your den'rs of the terms.

Your terms are 1/4, 1/5, 1/6..... 1/9, right?

so just let n be your den'r:

\[\Large \sum_{n=4}^{9}\dfrac{1}{n}=\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{9}\]

So that is ALMOST it!! the only problem is, THAT^^^ is all sum, and what you want is to sum the terms where n is even, and SUBTRACT the terms where n is odd. That's where the use of \(\large (-1)^n\) comes in.

Do you see why \(\large (-1)^n=1\) IF n is EVEN, and \(\large (-1)^n=-1\) IF n is ODD?

Answer 12

"What's wrong with sum n = 4 9 of (1/n)*(-1)^n"

Nothing! that's perfect - that's it! But I don't think you said that before.... did you? lol :)

Answer 13

Ok i get the n in the denominator part, but the equation confuses me because it has to be set, but how can it change for the odd/even? or do i put both to cancel?

Answer 14

That's why you multiply by (-1)^n

Answer 15

so (-1)^n(1/n)?

Answer 16

Hmmm.... I'm not sure what you mean.... the only part where odd/even is relevant is for the (-1)^n.

But you don't have to DO anything - as long as you have that factor in each term, then the terms for n's that are EVEN will be ADDED, and the terms for n's that are ODD will be SUBTRACTED.

The (-1)^n takes care of that.

That's why I asked above: Do you see why \(\large (−1)^n=1\) IF n is EVEN, and \(\large (−1)^n=-1\) IF n is ODD?

Answer 17

yes, (-1)^n*(1/n) is what you are summing, just put it in the summation notation.

You could even write it as: [(-1)^n]/n but either way is correct, they are equivalent, whether you put the (-1)^n in the num'r, as as a "coefficient" out in front is entirely up to you.

Answer 18

So the proper stigma is [(-1)^n]/n or what i had?

Answer 19

Either. they are the same.

\(\Large (-1)^n\cdot\dfrac{1}{n}=\dfrac{(-1)^n}{n}\)

The are exactly the same thing.

But it's "sigma" not "stigma". There is no stigmatizing in math! :) :)

Answer 20

ok thank you very much. i have 1 more im stuck on ill post thats alike

Answer 21

you're welcome. :)