Question

Needing assistance with sigma problems. I need to see all steps. Will give medal to best answer.
I'll post a pic of the problems in the comments.

Answer 1

Think: is that final "1" to be added 6 times to n^2 for n={1,2,3,4,5,6}, or
are we just adding up n^2 for n={1,2,3,4,5,6} and adding 1 to the resulting sum? Let's be certain regarding which one is meant.

Answer 2

Well, my book says to use the terms after the sigma to solve the problem. So I plugged in 1. But wasnt sure where to go next.
What I had; (1)^2+1=2

Answer 3

There are several summation formulas to learn and apply.

\[\sum_{1}^{n}1=n\]
Are you familiar with the next one?

Answer 4

The very first formula is used when you're just summing up repeated occurrences of " 1 " .
The second is used when you're summing up repeated occurrences of i.
The third is used when you're summing up repeated occurrences of i^2.
Have you encountered and/or used these formulas before?

Answer 5

I just started doing problems like these so Im not very comfortable with them. :/ I dont think I am familiar with the ones you're talking about yet.

Answer 6

You have a choice of looking up "summation formulas" and using them, or you could simply write out values of i^2 for i:{1,2,3,4,5,6} and then sum up the results. Any preference?

Answer 7

Please see

http://tutorial.math.lamar.edu/Classes/CalcI/SummationNotation.aspx

and look under Formulas for the formulas in question. You might want to copy down the first three of these, as you'll likely need them later on.

Answer 8

If you decide to skip use of the formulas, you could do the first problem in the following manner: Add up 1^2, 2^2, 3^2, 4^2, 5^2, 6^2. Awaiting your decision and action.

Answer 9

Sarah, want to continue with this discussion or not?

Answer 10

Sorry @mathmale. I fell asleep last night. @surjithayer @Whiterabbit541

Answer 11

Good morning!

Please take a moment to review what we typed last night. Then please add up 1^2, 2^2, 3^2, 4^2, 5^2, 6^2.

Answer 12

I looked over that sight and Im not quite understanding. Which formula do I use for adding up 1^2 through 6^2? I know adding them works but I think I need to show little more in depth work to receive credit. I thought I would just have to plug in each number to the n^2+1?

Answer 13

Actually, I'm suggesting that you "plug each value of n, from 1 to 6, into the n^2+1.

Once you've done that, we'l use the formula method to obtain the same result.

Please see

http://tutorial.math.lamar.edu/Classes/CalcI/SummationNotation.aspx

and look under Formulas for the formulas in question. You might want to copy down the first three of these, as you'll likely need them later on.

Answer 14

I have
(1)^2+1=2
(2)^2+1=5
(3)^2+1=10
(4)^2+1=17
(5)^2+1=26
(6)^2+1=37

Answer 15

Would I use the second formula on that section of the page?

Answer 16

First: In response to your calculations: Were you to add up the sums 2, 5, 10, 17, 26 and 37, you'd have the answer to the first question posed here.

Answer 17

Oh! I was over thinking it I guess. So the sum of the series is 97?

Answer 18

Responding to y our question about which formula to use: Mind explaining why you've focused on Formula 2 first?

Answer 19

Yes...that sum of 97 looks reasonable (although I'm not actually doing the math myself).

Answer 20

It seems to match up closest to the problem I have.

Answer 21

That problem involves n^2, not n, so wouldn't Formula #3 be more appropriate?

Answer 22

Oh, yeah I guess so. I was paying more attention to the formulas next to the sigmas. So for my second problem, would I just plug in 0 through 5 to the 16(1/4)^n as n?

Answer 23

Or is that not all of the work needed for the first one?

Answer 24

I think it best that we finish the first problem before tackling the second, but that decision is up to you. My point is that you can just add up the squares of the numbers 1 thru 6, obtaining 97, or you could use Formula #3. It'd be best if you could do this problem in two different ways: direct calculation and via formula #3.

Answer 25

Okay, Ill try using Formula #3. Do I just plug in each value for n? Or just the highest value, 6?

Answer 26

Just the highest value, 6.

Answer 27

If you focus only on the n^2 (not on the 1), you should get 91 from Formula #3.

Answer 28

I just finished it and I did get 91. So what is the right answer then? If the formula says 91 but direct calculation says 97.

Answer 29

Right, because you applied Formula #3 to ONLY n^2.
Apply Formula #1 to that 1, and you'll get 6.
Then, obviously, 91+6=97.

Answer 30

You're doing fine, Sarah!

I need to get off my computer now. However, you could continue posting questions here on OpenStudy in the usual way, or you could send me questions via Messaging to which I'd respond later on. For now, take care, keep up the good work!!

Answer 31

Okay, thank you! I'm gonna take a shot at the next problem & post it for someone to check. I appreciate all of your help!

Answer 32

I don't expect to be on my computer for long, but did want to help you get started on this new problem.

This expression is a GEOMETRIC SERIES. Having a name to look up may help.
I note that when n=0, 16(1/4)^n = 16; we call that the first term of the series, and denote it by "a".

That (1/4) is the COMMON RATIO, usuallly denoted by "r".

The sum of an infinite geom. series, provided that certain criteria are met, is simply

a
--------
1 - r

What is that criterion/criteria?

This particular problem is not an infinite geom. series, as n goes only from 0 to 5.

There is another formula for the sum of a geom. series when the series is not an infinite one. Can you find that formula through some research (in textbook, notes, online or whatever)?

Answer 33

\[\sum_{1}^{6}n ^{2}+1\]
\[we~know~\sum n ^{2}=\frac{ n \left( n+1 \right)\left( 2n+1 \right) }{6 }\]
put n=6
\[\sum_{1}^{6}n ^{2}+1=\frac{ 6\left( 6+1 \right)\left( 2*6+1 \right) }{ 6 }+1=\frac{ 6*7*13 }{ 6 }+1=91+1=92\]

Answer 34

Surji: Thanks for this different perspective. Strictly speaking, your interpretation of this problem appears to be correct. Sarah's interpretation ("Well, my book says to use the terms after the sigma to solve the problem. So I plugged in 1." sounded as tho she was to proceed as tho' both n^2 and 1 were within parentheses: (n^2 + 1).

Answer 35

@Sarah1017 : If I had to choose, I'd go with Surji's assumption that that " 1 " is added on to the sum of the first six terms of n^2. Might be a good idea to discuss these interpretations with your teacher, so you'll know better what to do on a test.