Question

Write the following sums in summation notation...

Answer 1

\[-\frac{ 3 }{ 2^{2} } +\frac{ 5 }{ 3^{2} }-\frac{ 7 }{ 4^{2} }+\frac{ 9 }{ 5^{2} }- \frac{ 11 }{ 6^{2} }+\frac{ 13 }{ 7^{2} }\]

Answer 2

Can you explain step by step please

Answer 3

step #1
you are alternating signs, so you will have as a factor in each term:
(-1)^(even or odd power)
so that can be controlled by your index.

step #2
your numerator is odd and separated by 2. That can also be controlled
by your index by using a multiple of 2 and adding "1" to get
successively increasing odd numbers.

step #3
your denominator is increasing by 1 but is 1 greater than your index.
again, that is controllable by your index by adding 1 each time to what your
index is.

Answer 4

So I have to come up with an equation that will equal to these sums?

Answer 5

What you have is an expression not an equation because there is no "=" sign. You will need to use sigma notation:\[\sum_{k=1}^{6}\]because you have 6 terms, but it is not set equal to anything at this time.

Answer 6

You don't have to use "k" specifically, but "k" is commonly used. Other common indices are "n" and "x". The "6" is for 6 terms.

Answer 7

And btw, the sigma is from using the equation tool. You might want to hit the "equation" button at the bottom to see it.

Answer 8

If you are having trouble with the tool, that's fine, you can skip that part, because that's done for you already anyway (above). You can now just concentrate on the term written in "k" that comes after the sigma.

Answer 9

\[\sigma ???\]

Answer 10

I'm so confused where did sigma come from? :S

Answer 11

Look at my second post for the sigma summation notation. That's capital sigma and it stands for "adding up all terms as k goes from 1 to 6". If your teacher wants you to use summation notation, then I am sure he introduced the symbol already and how to use it.

Answer 12

Oh yes you mean the big e looking thing lol

Answer 13

So I have to come up with the rest of the expression

Answer 14

Yes, you have to come up with the rest based on my first post. It's all there in detail. Try something and show some work so I'll know where you are in your understanding.

Answer 15

Like I really just don't understand and he never gave us examples to look at to even try to understand it

Answer 16

Here's a hint: since "k" goes from 1 to 6, the number "1" corresponds to the first term for the value of "k". For the negative/positive factor, you are using (-1)^(some power). As "k" goes from 1 to 6, it alternates even and odd. You can take advantage of that property and use some suitable expression using "k" for the exponent for this factor. That's a pretty big hint and you should be able to get the alternating sign factor from this even more detailed description.

If he didn't give examples, then I can give you some examples here, but not just the answer to your problem. You have to do the work at Openstudy. We just guide you and you're getting quite a bit here. If your teacher is not giving you examples, then confront him.

Answer 17

I'm not asking for you to do the work, I want to understand it. You're throwing terms at me and i can't help if I don't understand what you're talking about lol. Give me another minute.

Answer 18

Here's some examples:

2 + 4 + 6 + 8 + 10 is \[\sum_{k=1}^{5} 2k\]

Answer 19

Here's another:

1 + 2 + 6 + 24 + 120 is\[\sum_{k=1}^{5} k!\]

Answer 20

so it's basically a formula made up to express the numbers? omg why a !???

Answer 21

wait I think I'm on to something lol

Answer 22

Here's another:

5 + 7 + 9 + 11 is\[\sum_{k=1}^{4} (2k + 3)\]

For the previous one, "5!" means 5 x 4 x 3 x 2 x 1. It's the number times all integers decreasing until you get to "1".

Answer 23

\[\sum_{k=1}^{6} \frac{ 3 }{ 2k^{2} } ??\]

Answer 24

5 + 7 + 9 + 11

can also bewritten with the index starting at "0"\[\sum_{k=0}^{3} (2k + 5)\]

Answer 25

was I even anywhere close?

Answer 26

is there some kind of formula I'm missing out on?

Answer 27

You can always check your answer by expanding it. Your summation expands to:

3/2 + 3/8 + 3/18 + 3/32 + 3/50 + 3/72

which does not match the sequence given.

There is no formula you are missing out on. You yourself are the formula-generator.

Answer 28

Notice how your terms are all positive, you'll want to address that by re-reading my post #1, step #1.

Notice how all your numerators are 3, you'll have to review step #2.

And finally, the denominators are off, so review step #3.

Answer 29

what about \[\sum_{k=1}^{6} -\frac{ 3+2k }{ k+2^{2} }\]

Answer 30

2k+1? for numerator

Answer 31

It's like the old expression: "How do you eat an elephant?" The answer is is "One bite at a time". I suggest breaking the problem down into components. Just start with step #1 and get the (-1)^(some power) factor. You need to alternate the signs. You know you have to have this factor. Just start there.

I'll make it easy for you: Try this problem and it will help you a LOT:

Put the following into summation notation:

1 - 1 + 1 - 1 + 1

It's a trivial type of problem, but it will organize your approach.

Answer 32

2k + 1 for the numerator is VERY close, but you still have to consider the alternating signs. So, take that into account.

Answer 33

okay hmm for that one would it be k^-1? so i put it to a negative power? to make it alternate

Answer 34

Think of 2k + 1 as PART of the numerator. The numerator will be a factor of (2k + 1) and that (-1)^(some power) factor. Then you will have the numerator nailed down. You're not reading my suggestions. In each suggestions is a partial answer. Just build off of it.

Answer 35

2k^-1 +1

Answer 36

?

Answer 37

for the numerator :(

Answer 38

Did you read the part where I said that the numerator contains 2 factors? Please read my suggestions. Each time I give one, I am giving not only a suggestion but a partial answer that you can build on.

2 factors in the numerator.

again, 2 factors in the numerator.

One factor is (2k + 1)

So, the numerator is going to be THAT times (-1)^(some expression in k).

Answer 39

All you have left to do for the numerator is figure out what "some expression in k" is.

Answer 40

(2k+1)(-1^K) just a guess... what do you mean some expression in k? k=1??

Answer 41

sorry I'm stupid... lol :\

Answer 42

or do you mean like it would be -1^-k

Answer 43

You are close. It's not

(2k + 1)(-1^k)

It's

(2k + 1)(-1)^k

It is VERY VERY VERY VERY VERY VERY important where parentheses go. Your expression does the exponent on "1" first and then adds the "-" sign last, after eponentiation. That will never do.

Answer 44

oh true. so that'll make it alternate from negative to positive

Answer 45

If you write it correctly like I showed you, then it will alternate. Okay, you have the numerator, all you have left is the denominator. We are done with 2 of the 3 steps.

Answer 46

just so I can look at it better
step #3
your denominator is increasing by 1 but is 1 greater than your index.
again, that is controllable by your index by adding 1 each time to what your
index is

Answer 47

(2k+1)(k^2)??? close?

Answer 48

Yes, so in the first term, where the denominator is "2" when k is 1, we can express that 2 in terms of "k".

So for the first several terms: (disregarding the exponent for now)

2 when k is 1
3 when k is 2
4 when k is 3
5 when k is 4

Can you see a relationship? Each time, number is one more than "k". Build from that.

Answer 49

So, your first attempt on the denominator is pretty far off, so use my last post to figure the denominator.

Answer 50

I actually gotta run to class there now but I'm going to take a look at this again after and hopefully I'll figure it out. Thanks for helping, I understand it a lot more than I did.

Answer 51

Hold on, the denominator is:

(k + 1)^2

Answer 52

You're welcome. Just go over it all and it will sink in.