# Graphing Calculator Question : Sigma Notation



## jaybird0827

Anyone here have a TI-83 or TI-84 graphing calculator?

I'm trying to find a way to do "Sigma Notation" sums. I would think there should be a way that we could pull up a Sigma, give it a function with an interval of values, run it and get the sum. 

Here's a simple example.

10
SIGMA n^2
n = 1

[Enter] should give me the sum total of 1 + 9 + 16 + ... + 100.

Is there something built-in, or do I need write and store a program in the calculator to do it?


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## Grymir

I have a TI-86, and it doesn't appear to do it exactly like you wanted. You could graph it using the graphing function, and then use the tracing feature to find the answers. But I would program it using a simple for next loop to print the results.


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## jwithnell

Hi,

I'm jwithnell's husband -- I'm a math teacher, and the ti-83 has a couple of functions that will allow you to do what you are asking.

The first is seq() and the other is sum(). You get to both through the 2nd LIST button: seq() through 2nd LIST OPS, sum through 2nd LIST MATH.

The seq function takes an expression, then a variable to increment, the start value, the end value and then an optional step size. What it generates is a list of values of the expression. So *seq(x^2, x, 1, 100, 1)* would generate the values 1, 4, 9, 16, ... 10000.

The second function, sum(), will then add the values in a list, so it becomes very easy to do the sigma through the two functions.

*sum(seq(x^2,x,1,100))*

which will then produce the final answer you want 338350. Of course for that particular series, there is a formula ... but I leave that as an exercise for the student! (hint: it is a cubic function)


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## Grymir

Cool! Thanks. I thought it was strange that the calculator couldn't do it.


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## Neopatriarch

jaybird0827 said:


> 10
> SIGMA n^2
> n = 1



I use a Voyage 200 now. If I need to find the sum from 1 to 11 of x^3, I enter:

sigma(x^3, x, 1, 11)

to get 4356. I think the TI-83 works the same.

For my problem, the summation formula is:

(n^2(n+1)^2)/4 with n=11


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## jwithnell

jwithnell's husband again.

If you want to use a different calculator, the TI-89 has an actual sigma (uses the greek sigma) that you enter the expression, variable, start, and stop values. You get to it through the calculus functions (f3 key in "normal" home). It will do the summation symbolically, as well as nearly any other math symbolically, so you can get a useful answer.

The 83 and 84 series you have to combine the functions I mention above.

If you have Microsoft Math, you can also enter the expression directly on the computer. I believe the TI-nspire also includes a sigma.


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## jaybird0827

I downloaded a program called SIGMA to my TI-84. It's kind of clunky because you have to press [enter] as many times as you have terms in the sequence. It's okay if you have like 4 or 5 terms or so.

-----Added 11/29/2008 at 03:59:29 EST-----

Using the TI-84, I followed the suggestion below; although I found that when I did SEQ first and then SUM I had to do SUM(2nd ANS) and that worked fine. I like it because I can compute the sum of the series in the second step rather than having to keep track of how many times I press [enter]. I like the "nested function" version even better that does it in one step, e.g. *sum(seq(x^2,x,1,100))*. Thank you jwithnell's husband!



jwithnell said:


> Hi,
> 
> I'm jwithnell's husband -- I'm a math teacher, and the ti-83 has a couple of functions that will allow you to do what you are asking.
> 
> The first is seq() and the other is sum(). You get to both through the 2nd LIST button: seq() through 2nd LIST OPS, sum through 2nd LIST MATH.
> 
> The seq function takes an expression, then a variable to increment, the start value, the end value and then an optional step size. What it generates is a list of values of the expression. So *seq(x^2, x, 1, 100, 1)* would generate the values 1, 4, 9, 16, ... 10000.
> 
> The second function, sum(), will then add the values in a list, so it becomes very easy to do the sigma through the two functions.
> 
> *sum(seq(x^2,x,1,100))*
> 
> which will then produce the final answer you want 338350. Of course for that particular series, there is a formula ... but I leave that as an exercise for the student! (hint: it is a cubic function)


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