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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) = 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 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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