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Homework Statement
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.
Homework Equations
##\sum\limits_{k=1}^n a_k= a_1 +a_2+a_3...+a_n##
The Attempt at a Solution
[/B]
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) = exponentgeneral 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 2x, 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
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