Sigma notation and calculator help(casio fx9860g; ti-30xpro)

1. The problem statement, all variables and given/known data
A) use sigma notation and calculate sum as far as you wish (set your own endpoint)


(1/2) + (1/4) + (1/8) + (1/16)...


B) this is a personal problem because I want to know how to calculate the sums with e.g. sigma notation, with a calculator.

Never having done this before, I have two calculators available. An older model graphing calculator casio fx-9860G, and a calculator Ti-30X Pro multiview. I was wondering if anybody knew how to input the conditions of the sigma, and get the correct type of calculation into the calculator.


I do not posess the manual book any longer, for casio fx-9860G but that particular calculator has some nostalgia value to me, and I know how to use it in many other situations.

2. Relevant equations

##\sum\limits_{k=1}^n a_k= a_1 +a_2+a_3....+a_n##

3. The attempt at a solution


A)

I was confused about the general form formula for sigma.

How can I make sure, that the "pattern variable" for the purpose of creating the n-th term of the series, is the correct variable from sigma notation itself?

in other words if the sigma starts such as ##\sum\limits_{k=1}^3 2^k##
does that equal to:
##2^1 + 2^2 + 2^3##

If instead I wrote sigma such as ##\sum\limits_{k=1}^6 2^(n-1)##
does that equal to:
##2^(6-1=5)+ 2^(5) + 2^5 +2^(5) +2^(5) +2^(5)##
because the series is limited from the first term, until the sixth term, where n=6 initially, but gets reduced by one when calculating? If the code looks bad it was supposed to be such that (n-1) = exponent


general observations about patterns
1. the top number stays as 1 in each term of the series (does it matter if you call sigma sum, as series, is there any difference between a sum and a series like in this example?). If I remember correctly, sequences are simply numbers which follow some pattern, but the numbers are simply divided by a comma, and no mathematical operations as such are performed after creation of the terms of the sequence.

2. the bottom number changes I think in the form of 2^x, where x is the index number of the term, which we are calculating. starting index being 1 for the first term.

such as the first term = ##\frac{1}{2^1}##
2nd. term= ##\frac{1}{2^2}##
3rd term = ##\frac{1}{2^3}##

This leads me to conclude thatthe sigma notation can now be created, because we know the pattern for creation of new terms for the series.

##\sum\limits_{i=1}^n \frac{1}{2^i} = ~~ \frac{1}{2^1}~~+ ~~ \frac{1}{2^2}~~+~~ \frac{1}{2^3}~~...##

the sum for the first 5 terms calculated manually = 0.5 + 0.25 +0.125 +(1/16) +(1//32)=31/32

 

ok i had a brain fart and the answer is 31/32 for the sum of 1-->5 terms

well ok I get now that n does not need to be the topcondition of the sigma.

however, if n were the top condition of the sigma, would that essentially be correct interpretation.

##\sum\limits_{k=1}^5 2^{n-1}= 2^4+2^4+2^4+2^4+2^4##

my textbook used the sigma notation such that
n= index of the upper bound term of the sum, included
k= index of the first term of the sum, included

 

my textbook shows
##\sum\limits_{k=m}^{n}a_k=a_m +a_{m+1}+...a_{n-1}+a_{n}##
where a_n = final term
and where a_m = the first term (ostensibly???)

with term of the sum being for example a_k = 2^k
Normally the sum would be something like
##\sum\limits_{k=1}^{5}2^k= 2^1+2^2+2^3+2^4+2^5##
Now if I wanted to calculate sum from the third term, until the fifth term (3rd + 4th + 5th)
If I wanted to calculate the sum from the 3rd term until the 5th term, could I do that with sigma notation (skip counting the 1st and the 2nd terms of the sum)

 

Well since you talked about being very specific with the n variable as e.g. the upperbound of the sum I could easily write on paper such that
##\sum\limits_{k=1}^{n=4}3^n=3^4+3^4+3^4+3^4##
Latex code can get a bit annoying at times, of course. Whether that above, is standard convention or not.

I think it would be accurate but a little bit silly example admittedly. If you can put k=1 on the bottom of the sigma why cant you put n=(whatever natural number) on the top?

but what about the next question.

##\sum\limits_{k=3}^{5}2^k= ## ##2^3+2^4+2^5## is this so???

 

I figured out on my own how to use sigma sums on my newer texas instruments Ti -30x pro calculator.

However I was unable sadly to understand how to calculate sums with casio fx-9860g