Question

Use summation nation to write each arithmetic series for the specified number of terms. Then evaluate the sum.

50 + 55 + 60 + ... ; n=7

What is summation nation? @mathmale @kirbykirby

Answer 1

I see that each subsequent term of this arith. series is 5 greater than the preceding term. Agree with that?

Answer 2

Since you need only 7 terms, you could obtain the remaining four in your head; the 4th is 65.

Answer 3

Yes and okay, how do I write out the answer? Does it have to be in a certain format?

Answer 4

Also, do they mean just in all or 7 after the 3 given?

Answer 5

*mean 7 in all

Answer 6

The first term is 50.
We add 5 to that to get 55.
The third term is 60, which is 55 + 5.
Please try to predict the fourth thru 7th terms.
All you need are the first 7 terms.

Later the problem becomes a bit more complicated: we have to use summation notation to predict the nth sum.

Answer 7

Okay, 50, 55, 60, 65, 70, 75, 80
And this one says to use summation nation.. Do I just write 50, 55, 60, 65, 70, 75, 80 on my page?

Answer 8

Yes, although that would not be your final answer to this question. I thought it'd be helpful to write out these 7 terms so that you could add them up manually and thus know if the rest of your work on this problem is correct or not.

Answer 9

So I don't need to write anything in summation nation?

Answer 10

Here's what I would do, towards finding a formula for the nth term:

50+0(5) = 50 (first term)
50+1(5) = 55 (second term)
50+2(5) = 60 (third term)
50 +3(5) =65 (fourth term)
If you see my pattern, continue my pattern to find the fifth term, please.

Answer 11

50 + 4(5) = 70
50 + 5(5) = 75
50 + 6(5) = 80
50 + 7(5) = 85

Answer 12

Opps just to 80

Answer 13

Right. But that's OK. Recognizing the pattern is the key issue here. Nice work!

Now let's change our findings into a formula for the nth term of this arith. sequence, OK?

Answer 14

Okay, no idea how to do that ha

Answer 15

the first term is 50. I wrote that as 50 + 0(5).
Note that n = 1 signifies "first term."
Note that that 0 can be written as 1-1.

Thus, I propose that the formula for the first term is 50 + (1-1)(5).
Does this produce the expected result, 50?

Answer 16

in other words, Nicole, does 50 + (1-1)5 = 50 + 0(5)?

Answer 17

Yes, it dose. So do I replace the 1-1 with "n" to signifies the replacement of the number for the equation?

Answer 18

Exactly! would you do that, please?

The formula for the nth term is ??

Answer 19

50 + n(5) = An ?

Answer 20

think carefully about what you are replacing with n. We don't replace all of that "1-1" with n.

Test your theory. your formula says that the nth term is 50 + n(5).
Let's see whether that predicts the first term properly.

Let n = 1 in your equation. What is the result?

Answer 21

oh, not 50..
Umm, 50 + n-1(5) = An ?

Answer 22

If only you'll enclose that "n-1" inside parentheses, YES!

Please do that.

The formula for the nth term is ... ??

Answer 23

50 + (n-1)5 = An

Answer 24

Or 50 + (n-1)(5) = An ?

Answer 25

Wait, no I take that ^^ back..

Answer 26

Let's see what kirbykirby has to say!

Answer 27

Okay :P

Answer 28

In the meantime: What leads you to "take that back?"

Answer 29

Hm well I just wanted to add that the question asks for a series, so shouldn't we be adding the terms, and thus use sigma notation, rather than just a sequence. Or maybe you are getting to that, which doesn't take much effort lol. But good job on the explanation so far mathmale

Answer 30

FIOL I realized I would change the out come..

Answer 31

I do believe we are 1) expressing a series in summation form, and 2) finding the actual sum of the series when n=7.

Answer 32

I think so.. The very start of this post has everything I'm to do..

Answer 33

Nicole, have you seen these formulas before?

Answer 34

Parts look familiar but I didn't even know what summation nation was when we started this problem.. I don't think I could explain it now either

Answer 35

\[\sum_{i-1}^{n}1=n;\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }??\] ... these are the formulas I was referring to.

Answer 36

No.. I don't know

Answer 37

The first formula says that if we add up n 1's, the sum is just n.

Answer 38

Okay

Answer 39

Will you have any contact with your teacher tomorrow?
If so, i advise that you ask wehther or not you're expected to learn and know these particular summation formulas.

Answer 40

I forget what the \[\sum_{ }^{ }\]Means..

Answer 41

I believe there's another formula

Answer 42

I am, I just have a problem remembering and my teacher doesn't explain it in a way that I can easily pick up, where most of the time you guys and my tutor does..

Answer 43

Oh, thats a bit confusing

Answer 44

\[\sum_{i=1}^{n}1=n=1+1 + 1 + 1 + 1 + ....\] until you have added n 1's together.

Answer 45

Yes, I made a mistake, which has now been corrected.

Answer 46

\[\sum_{i=1}^{3}1=1+1+1=3=n\]is an easy example.

Answer 47

Ohh, okay. I get it now.. I think ha

Answer 48

I won't be able to provide a proof, but the following is also true:\[\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }\]

Answer 49

Again, Nicole, as soon as you can, contact your teacher and find out if and when you need to know these "summation formulas". I would bet you do.

Answer 50

So for my problem all that needs to be on the page is this right?

50 + 55 + 60 + 65 + 70 + 75 + 80


50 +0(5) = 50
...
50 + 1(5) = 55


50 + (n-1)5 = An

Answer 51

^Proof by induction :) haha but I think that would be too much lool

Answer 52

Opps

50 +0(5) = 50
...
50 + 6(5) = 80

Answer 53

I thought it was \[\sum_{k = 0}^{n-1} (a+kd) = \frac{ n }{ 2 }(2a(n-1)d)\]

Answer 54

Oh I never remember such specific formulas :S Only the basic ones needed

Answer 55

typo, (2a + d(n-1))

Answer 56

I believe there are different versions of the same general formula. I'm merely presenting the version with which I'm most familiar.

Answer 57

Is that right?

Answer 58

Its really a matter of plugging in the values you already know, a, d and n

Answer 59

I will let The Fizic answer that. He will first have to define what he means by a, d and n.

Answer 60

So I need more for my page?

Answer 61

Ladies and gentlemen, as much as I'd like to stick with this discussion, I need to hit the sack. Been on Open Study approx ten hours today.

Answer 62

oh my @mathmale

Answer 63

Okay, Thank you MM!

Answer 64

So I bid you both a fond "good night." kirbykirby too!

Answer 65

Good night!!

Answer 66

\[\sum_{k = 0}^{7-1}(50 + k * 5) = 7/2(2(50) + 5(7-1))\]

Answer 67

@kirbykirby Is this all I need for my answer?

50 + 55 + 60 + 65 + 70 + 75 + 80


50 +0(5) = 50
...
50 + 6(5) = 80


50 + (n-1)5 = An

Answer 68

Nicole, yes a bit more is needed for your answer. Also, the question specifically asks for 7 terms.

Answer 69

Buenas Noches, Luis! (Hey, that was an interesting expression; too bad you've erased it.

Answer 70

Yes Fizi has the right formula down

Answer 71

Over and out! Bye!

Answer 72

There is 7, 50 to 80.

Answer 73

I did? goodnight Warren!!

Answer 74

\[\sum_{i=0}^6 (50+5i)\]

Answer 75

you can replace "i" with "k", it doesn't matter what letter is used.

Answer 76

I am so confused.. I have 5 other problem I don't know how to do. So I think I'll just leave with one with what I have. If I only get partial point on the problem then so be it haha

Answer 77

xD

Answer 78

do you know how to evaluate something like this:
\[\sum_{i=1}^4 i\]

Answer 79

I don't know what they E/Z thing is..

Answer 80

*the

Answer 81

it is a summation operator. It is basically a short form to write out a sum.
So the sum I wrote above, it just saying "Add all the terms going from 1 to 4, by replacing "i" with 1, 2, 3, 4"

Answer 82

Maybe some examples will help:

Answer 83

\[\sum_{i=1}^4 i=1+2+3+4\]\[\sum_{i=1}^6 i=1+2+3+4+5+6\]\[\sum_{i=3}^7 i=3+4+5+6+7\]

Answer 84

So more complicated expressions now:
\[\sum_{i=1}^4 2i=2(1)+2(2)+2(3)+4(2)\]\[\sum_{i=1}^4 (5+i)=(5+1)+(5+2)+(5+3)+(5+4)\]

Answer 85

Now, there is a formula, that mathmale mentioned:
\[\sum_{i=1}^n i=\frac{n(n+1)}{2}\]. Why do we need this? Well, imagine if "n" was 100. We certainly would not want to add each number individually up to 100 (much too long to do so). So the formula above will gove you the sum of numbers from 1 to 100, if you replace "n" by 100.

Answer 86

Okay, can you show me what mine would look like? @kirbykirby

Answer 87

\[\sum_{i=0}^6 (50+5i)\]
This is because it would give you these terms:
[50+5(0)] + [50+50(1)] + [50+50(2)] + [50+50(3)] + [50+50(4)] + [50+50(5)] + [50+50(6)]

Answer 88

even though it starts at 0 and ends at 6, please do realize that this gives you 7 terms. It might "look" unintuitive until you actually count the terms.

Answer 89

Okay, thank you! I think I get it but I'm not sure. I have another problem with a different sequence, could you help me with that one as well?

Answer 90

okay

Answer 91

21 + 19 + 17 + ... ; n = 8

I know its by twos

Answer 92

So, the first term... is 21
To get the 2nd term, using the first term, you do 21 - 2
Now, what do you do to get the 3rd term, 17

Answer 93

21, 19, 17, 15, 13, 11, 9, 7

Answer 94

Yes. But, we will want to write this in summation form. You want to construct ever number in your sequence based on the FIRST term. Like, for the previous question, we found every term of the sequence based on the first the 50, and see that we could get the n-th term by doing 50 + 5(n-1) (if n starts at 0)

Answer 95

21 - 0(2) = 21 ?

Answer 96

yes for the 1st term

Answer 97

21 - 1(2) = 19
21 - 2(2) = 17
21 - 3(2) = 15
21 - 4(2) = 13
21 - 5(2) = 11
21 - 6(2) = 9
21 - 7(2) = 7